PKU《程序设计导引及在线实践》刷题记录（上）
2024-04-30 05:41:41
2.1 鸡兔同笼
#include <cstdio>int main() {int n;scanf("%d", &n);while (n--) {int total;int chi, rab;scanf("%d", &total);if (total % 2 != 0) {printf("0 0\n");continue;}elseprintf("%d %d\n", total / 4 + (total - total / 4 * 4) / 2, total / 2);}return 0;
}
2.2棋盘上的距离
#include <cstdio>
#include <cmath>int main() {int ncases;scanf("%d", &ncases);while (ncases--) {int dx, dy;char x[5], y[5];scanf("%s %s", x, y);dx = abs(y[0] - x[0]);dy = abs(y[1] - x[1]);//王printf("%d ", dx > dy ? dx : dy);//后if (dx == 0 || dy == 0 || dx == dy)printf("1 ");elseprintf("2 ");//车if (dx == 0 || dy == 0)printf("1 ");elseprintf("2 ");//象if (dx != dy)printf("Inf\n");elseprintf("1\n");}return 0;
}
2.3 校门外的树
#include <cstdio>
#include <algorithm>
using namespace std;int tree[10010];int main() {int L, M;scanf("%d %d", &L, &M);fill(tree, tree + L + 1, 1);for (int i = 0; i < M; i++) {int begin, end;scanf("%d %d", &begin, &end);fill(tree + begin, tree + end + 1, 0);}int ans = 0;for (int i = 0; i <= L; i++)ans += tree[i];printf("%d", ans);return 0;
}
2.4填词
#include <cstdio>int main() {char c;int num[30] = { 0 };int M, N, P;scanf("%d %d %d", &M, &N, &P);getchar();for (int i = 0; i < M; i++) {for (int j = 0; j < N; j++) {scanf("%c", &c);num[c - 'A']++;}getchar();}for (int i = 0; i < P; i++) {char temp[500];scanf("%s", temp);for (int j = 0; temp[j] != '\0'; j++)num[temp[j] - 'A']--;}for (int i = 0; i < 26; i++) {while (num[i] != 0) {printf("%c", i + 'A');num[i]--;}}return 0;
}
2.5装箱问题
太菜了，没想到这题居然还是不会，没学好啊
#include <cstdio>int main() {while (1) {int ans = 0;int a, b, c, d, e, f;//用来存放订单数int x, y;//x表示剩下2*2的空间的数量，y表示剩下1*1的空间的数量scanf("%d %d %d %d %d %d", &a, &b, &c, &d, &e, &f);if (a == 0 && b == 0 && c == 0 && d == 0 && e == 0 && f == 0)break;int left[4] = { 0,5,3,1 };ans = f + e + d + (c + 3) / 4;x = 5 * d + left[c % 4];if (b > x)ans += (b - x + 8) / 9;x = 36 * ans - 36 * f - 25 * e - 16 * d - 9 * c - 4 * b;if (a > x)ans += (a - x + 35) / 36;printf("%d\n", ans);}return 0;
}
3.1确定进制
太菜了！！怎么感觉自己还不如第一遍刷题的自己？？居然忘了进制转换的时候，如果有个要被转换的数比product大，应当退出！！
#include <cstdio>int Change(char p[], int product) {int ans = 0;for (int i = 0; p[i] != '\0'; i++) {if (p[i] - '0' >= product)return -1;ans = ans * product + p[i] - '0';}return ans;
}int main() {char p[10], q[10], r[10];scanf("%s %s %s", p, q, r);for (int i = 2; i <= 16; i++) {int a = Change(p, i);int b = Change(q, i);int c = Change(r, i);if (a*b == c) {printf("%d", i);return 0;}}printf("0");return 0;
}
3.2相邻数字的基数不等比：skew数
#include <cstdio>
#include <cstring>int main() {int base[32];base[0] = 1;for (int i = 1; i < 32; i++)base[i] = 2 * base[i - 1] + 1;while (1) {char p[40];scanf("%s", p);if (p[0] == '0')break;long long ans = 0;int len = strlen(p);for (int i = len - 1; i >= 0; i--) {ans += (p[i] - '0')*base[len - i - 1];}printf("%ld\n", ans);}return 0;
}
4.2统计字符数
#include <cstdio>
#include <cstring>int main() {int n;scanf("%d", &n);while (n--) {char str[1010];scanf("%s", str);int num[30] = { 0 };int MAXID = 0;for (int i = 0; str[i] != '\0'; i++) num[str[i] - 'a']++;for (int i = 1; i < 26; i++) {if (num[i] > num[MAXID])MAXID = i;}printf("%c %d\n", 'a' + MAXID, num[MAXID]);}return 0;
}
4.3 487-3279
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
using namespace std;int main() {map<string, int> mp;int n;scanf("%d", &n);while (n--) {string s;cin >> s;string temp;//输入字符串并进行映射for (int i = 0; s[i] != '\0'; i++) {if (s[i] == 'A' || s[i] == 'B' || s[i] == 'C') {temp += "2";}if (s[i] == 'D' || s[i] == 'E' || s[i] == 'F') {temp += "3";}if (s[i] == 'G' || s[i] == 'H' || s[i] == 'I') {temp += "4";}if (s[i] == 'J' || s[i] == 'K' || s[i] == 'L') {temp += "5";}if (s[i] == 'M' || s[i] == 'N' || s[i] == 'O') {temp += "6";}if (s[i] == 'P' || s[i] == 'R' || s[i] == 'S') {temp += "7";}if (s[i] == 'T' || s[i] == 'U' || s[i] == 'V') {temp += "8";}if (s[i] == 'W' || s[i] == 'X' || s[i] == 'Y') {temp += "9";}if (s[i] >= '0'&&s[i] <= '9')temp += s[i];}temp.insert(temp.begin() + 3, '-');if (mp.count(temp) != 0)mp[temp]++;elsemp[temp] = 1;}bool flag = false;for (map<string, int>::iterator it = mp.begin(); it != mp.end(); it++) {if (it->second > 1) {cout << it->first << " " << it->second << endl;flag = true;}}if (flag == false)cout << "No duplicates.";return 0;
}
4.4子串
#include <iostream>
#include <string>
using namespace std;int main() {int n;scanf("%d", &n);while (n--) {int t;cin >> t;string str[110];string min;cin >> str[0];min = str[0];for (int i = 1; i < t; i++) {cin >> str[i];if (str[i].length() < min.length())min = str[i];}int minlen = min.length();int i;bool flag = false;for (i = minlen; i > 0; i--) {//子串长度从最长开始逐次递减for (int j = 0; j + i <= minlen; j++) {//子串起始位置枚举string temp;temp.insert(temp.begin(), min.begin() + j, min.begin() + i + j);flag = true;for (int k = 0; k < t; k++) {if (str[k] == min)continue;string rev;for (int s = str[k].length() - 1; s >= 0; s--)rev += str[k][s];if (str[k].find(temp) == string::npos&&rev.find(temp) == string::npos) {flag = false;}}if (flag == true)break;}if (flag == true) {printf("%d\n", i);break;}}if (flag == false)printf("0\n");}return 0;
}
4.5最难的问题
#include <iostream>
#include <string>
using namespace std;int main() {while (1) {string start, end, sent;getline(cin, start);if (start == "ENDOFINPUT")break;getline(cin, sent);getline(cin, end);string ans;for (int i = 0; sent[i] != '\0'; i++) {if (sent[i] >= 'A'&&sent[i] <= 'Z')ans += (sent[i]-'A' + 26 - 5) % 26 + 'A';elseans += sent[i];}cout << ans << endl;}return 0;
}
5.1判断闰年
#include <cstdio>int main() {int year;scanf("%d", &year);if (year % 4 != 0)printf("N");else {if (year % 3200 == 0) {printf("N");return 0;}if (year % 100 == 0)if (year % 400 != 0) {printf("N");return 0;}printf("Y");}return 0;
}
5.2细菌繁殖
#include <cstdio>int main() {int month[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };int n;scanf("%d", &n);while (n--) {int Mst, Dst, Men, Den;long long num;scanf("%d %d %lld %d %d", &Mst, &Dst, &num, &Men, &Den);int lag = 0;while (Mst != Men || Dst != Den) {lag++;if (Dst == month[Mst]) {Dst = 1;Mst++;}elseDst++;}for (int i = 0; i < lag; i++)num += num;printf("%lld\n", num);}return 0;
}
5.3 日历问题
#include <cstdio>
int month[2][13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,30,31 };
char week[7][10] = { "Saturday","Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" };
int year[2] = { 365,366 };bool IsRun(int year) {if (year % 4 != 0)return false;if (year % 100 == 0 && year % 400 != 0)return false;return true;
}int main() {while (1) {int lag;scanf("%d", &lag);if (lag == -1)break;int temp = 0;int Y = 2000, M = 1, D = 1;while (temp != lag) {bool flag = IsRun(Y);if (lag - temp > year[flag]) {temp += year[flag];Y++;}else {temp++;if (D == month[flag][M]) {D = 1;M++;if (M == 13) {M = 1;Y++;}}elseD++;}}printf("%04d-%02d-%02d %s\n", Y, M, D, week[lag % 7]);}return 0;
}
5.4玛雅历
这题一开始就不会做=-=，现在看到这个题还是犯怵。。。太菜了
#include <cstdio>
#include <cstring>char maya[19][10] = { "pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};
char holly[20][10] = { "imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau" };int main() {int n;scanf("%d", &n);printf("%d\n", n);while (n--) {int day, year;char mon[10];int sum = 0;scanf("%d. %s %d", &day, mon, &year);sum += year * 365;for (int i = 0; i < 19; i++) {if (strcmp(maya[i], mon) == 0) {sum += i * 20;break;}}sum += day;printf("%d %s %d\n", sum % 13 + 1, holly[sum % 20], sum / 260);}return 0;
}
5.5 时区间时间的转换
#include <cstdio>
#include <string.h>int difference(char zone1[], char zone2[]) {char zone[32][7] = {"UTC","GMT","BST","IST","WET","WEST","CET","CEST","EET","EEST","MSK","MSD","AST","ADT","NST","NDT","EST","EDT","CST","CDT","MST","MDT","PST","PDT","HST","AKST","AKDT","AEST","AEDT","ACST","ACDT","AWST" };double time[32] = { 0,0,1,1,0,1,1,2,2,3,3,4,-4,-3,-3.5,-2.5,-5,-4,-6,-5,-7,-6,-8,-7,-10,-9,-8,10,11,9.5,10.5,8 };int i, j;for (i = 0; strcmp(zone[i], zone1) != 0; i++);for (j = 0; strcmp(zone[j], zone2) != 0; j++);return (int)((time[i] - time[j]) * 60);
}int main() {int n;scanf("%d", &n);while (n--) {char time[9];int hours, minutes;scanf("%s", time);switch (time[0]) {case'n':hours = 12;minutes = 0;break;case'm':hours = 0;minutes = 0;break;default:sscanf(time, "%d:%d", &hours, &minutes);hours %= 12;scanf("%s", time);if (time[0] == 'p')hours += 12;}char timezone1[7], timezone2[7];scanf("%s %s", timezone1, timezone2);int newtime;newtime = hours * 60 + minutes + difference(timezone2, timezone1);if (newtime < 0)newtime += 1440;newtime %= 1440;switch (newtime) {case 0:printf("midnight\n"); break;case 720:printf("noon\n"); break;default:hours = newtime / 60;minutes = newtime % 60;if (hours == 0)printf("12:%02d a.m.\n", minutes);else if (hours < 12)printf("%d:%02d a.m.\n", hours, minutes);else if (hours == 12)printf("12:%02d p.m.\n", minutes);else if (hours > 12)printf("%d:%02d p.m.\n", hours % 12, minutes);}}return 0;
}
6.1 约瑟夫问题
#include <cstdio>typedef struct Monkey {int data;struct Monkey *next;
}monkey, *linklist;int main() {while (1) {linklist Head;Head = new Monkey;Head->next = NULL;linklist Tail;Tail = Head;int num, k;scanf("%d %d", &num, &k);if (num == 0 && k == 0)break;if (k == 1)printf("%d\n", num);else {for (int i = 1; i <= num; i++) {linklist temp;temp = new Monkey;temp->data = i;temp->next = NULL;Tail->next = temp;Tail = temp;}Tail->next = Head->next;delete Head;linklist L = Tail->next;while (L->next != L) {for (int i = 1; i < k - 1; i++)L = L->next;linklist temp = L->next;L->next = temp->next;delete temp;L = L->next;}printf("%d\n", L->data);}}return 0;
}
6.2 花生问题
很奇怪，OJ仿佛识别不了math.h里的abs？
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;int Field[55][55];int main() {int T;scanf("%d", &T);while (T--) {int M, N, K;scanf("%d %d %d", &M, &N, &K);for (int i = 1; i <= M; i++)for (int j = 1; j <= N; j++)scanf("%d", &Field[i][j]);int Time = 0;int Sum = 0;int curi = 0, curj = 0;while (Time <= K) {//首先找到最大的int maxn = 0;int maxi = 0, maxj = 0;for (int i = 1; i <= M; i++)for (int j = 1; j <= N; j++) {if (Field[i][j] > maxn) {maxn = Field[i][j];maxi = i;maxj = j;}}if (maxn == 0)break;if (curi == 0)curj = maxj;if (Time + maxi + 1 + abs(maxi - curi) + abs(maxj - curj) <= K) {Time += 1 + abs(maxi - curi) + abs(maxj - curj);curi = maxi;curj = maxj;Sum += Field[maxi][maxj];Field[maxi][maxj] = 0;}elsebreak;}printf("%d\n", Sum);}return 0;
}
6.3显示器
主要方法：用char字符数组存每个位置都被哪个数字占了
#include <cstdio>
#include <string.h>
char n1[11] = { "- -- -----" };
char n2[11] = { "| ||| ||" };
char n3[11] = { "||||| |||" };
char n4[11] = { " ----- --" };
char n5[11] = { "| | | | " };
char n6[11] = { "|| |||||||" };
char n7[11] = { "- -- -- --" };
int main() {while (1) {int s;char szNumber[20];int nDigit, nLength, i, j, k;scanf("%d %s", &s, szNumber);if (s == 0)break;nLength = strlen(szNumber);for (int i = 0; i < nLength; i++) {nDigit = szNumber[i] - '0';printf(" ");for (j = 0; j < s; j++)printf("%c", n1[nDigit]);printf(" ");printf(" ");//这个是数字间的空格}printf("\n");for (i = 0; i < s; i++) {for (j = 0; j < nLength; j++) {nDigit = szNumber[j] - '0';printf("%c", n2[nDigit]);for (k = 0; k < s; k++)printf(" ");printf("%c", n3[nDigit]);printf(" ");}printf("\n");}for (int i = 0; i < nLength; i++) {nDigit = szNumber[i] - '0';printf(" ");for (j = 0; j < s; j++)printf("%c", n4[nDigit]);printf(" ");printf(" ");}printf("\n");for (i = 0; i < s; i++) {for (j = 0; j < nLength; j++) {nDigit = szNumber[j] - '0';printf("%c", n5[nDigit]);for (k = 0; k < s; k++)printf(" ");printf("%c", n6[nDigit]);printf(" ");}printf("\n");}for (int i = 0; i < nLength; i++) {nDigit = szNumber[i] - '0';printf(" ");for (j = 0; j < s; j++)printf("%c", n7[nDigit]);printf(" ");printf(" ");}printf("\n");printf("\n");}return 0;
}
6.4 排列
STL大法好
#include <cstdio>
#include <algorithm>
using namespace std;int main() {int n;scanf("%d", &n);while (n--) {int a[1050];int t, k;scanf("%d %d", &t, &k);for (int i = 0; i < t; i++)scanf("%d", &a[i]);for (int i = 0; i < k; i++)next_permutation(a, a + t);for (int i = 0; i < t; i++)printf("%d ", a[i]);printf("\n");}return 0;
}
7.1 大整数加法
我是猪……为啥做题做的还不如第一次做的好？？忽略了0+0的问题，WA好几次
#include <cstdio>
#include <string.h>int main() {char a[210], b[210];int ai[210] = { 0 }, bi[210] = { 0 };int ans[210] = { 0 };scanf("%s", a);scanf("%s", b);int alen = strlen(a), blen = strlen(b);for (int i = alen - 1, j = 0; i >= 0; i--, j++)ai[j] = a[i] - '0';for (int i = blen - 1, j = 0; i >= 0; i--, j++)bi[j] = b[i] - '0';int max = alen > blen ? alen : blen;int carry = 0;for (int i = 0; i < max; i++) {ans[i] = (ai[i] + bi[i] + carry) % 10;carry= (ai[i] + bi[i] + carry) / 10;}if (carry != 0)ans[max] = carry;while (ans[max] == 0)max--;if (max < 0)printf("0");else {for (int i = max; i >= 0; i--)printf("%d", ans[i]);}return 0;
}
7.2 大整数乘法
#include <cstdio>
#include <string.h>int main() {char a[210], b[210];int ai[210] = { 0 }, bi[210] = { 0 };int ans[450] = { 0 };scanf("%s", a);scanf("%s", b);int alen = strlen(a), blen = strlen(b);for (int i = alen - 1, j = 0; i >= 0; i--, j++)ai[j] = a[i] - '0';for (int i = blen - 1, j = 0; i >= 0; i--, j++)bi[j] = b[i] - '0';for (int i = 0; i < alen; i++)for (int j = 0; j < blen; j++)ans[i + j] += ai[i] * bi[j];for (int i = 0; i < alen + blen; i++) {if (ans[i] >= 10) {ans[i + 1] += ans[i] / 10;ans[i] %= 10;}}int max = alen + blen;while (ans[max] == 0)max--;if (max < 0)printf("0");else {for (int i = max; i >= 0; i--)printf("%d", ans[i]);}return 0;
}
7.3 大整数除法
我放弃，这题真要是考场上出来我可能会自杀
贴上书里给的代码
#include <cstdio>
#include <string.h>
const int MAXL = 200;
char szLine1[MAXL + 10];
char szLine2[MAXL + 10];
int an1[MAXL + 10];//被除数，an1[0]对应于个位
int an2[MAXL + 10];//除数，an2[0]对应于个位
int Result[MAXL + 10];//存放商，Result[0]对应于个位int Substract(int *p1, int *p2, int nLen1, int nLen2) {//判断p1是否比p2大，如果不是，返回-1if (nLen1 < nLen2)return -1;bool bLarger = false;if (nLen1 == nLen2) {for (int i = nLen1 - 1; i >= 0; i--) {if (p1[i] > p2[i])bLarger = true;else if (p1[i] < p2[i]) {if (!bLarger)return -1;}}}for (int i = 0; i < nLen1; i++) {p1[i] -= p2[i];if (p1[i] < 0) {p1[i] += 10;p1[i + 1]--;}}for (int i = nLen1 - 1; i >= 0; i--)if (p1[i])return i + 1;return 0;
}int main() {char szBlank[20];scanf("%s", szLine1);scanf("%s", szLine2);int nLen1 = strlen(szLine1);memset(an1, 0, sizeof(an1));memset(an2, 0, sizeof(an2));memset(Result, 0, sizeof(Result));for (int i = nLen1 - 1, j = 0; i >= 0; i--)an1[j++] = szLine1[i] - '0';int nLen2 = strlen(szLine2);for (int i = nLen2 - 1, j = 0; i >= 0; i--)an2[j++] = szLine2[i] - '0';if (nLen1 < nLen2) {printf("0");return 0;}nLen1 = Substract(an1, an2, nLen1, nLen2);if (nLen1 < 0) {printf("0");return 0;}else if (nLen1 == 0) {printf("1");return 0;}Result[0]++;int nTimes = nLen1 - nLen2;if (nTimes < 0)goto OutputResult;else if (nTimes > 0) {for (int i = nLen1 - 1; i >= 0; i--) {if (i >= nTimes)an2[i] = an2[i - nTimes];elsean2[i] = 0;}}nLen2 = nLen1;for (int j = 0; j <= nTimes; j++) {int nTmp;//一直减到不够减为止//先减去若干个an2*（10的nTimes次方）//不够减了，再减去若干个an2*（10的nTimes-1次方）while ((nTmp = Substract(an1, an2 + j, nLen1, nLen2 - j)) >= 0) {nLen1 = nTmp;Result[nTimes - j]++;//每成功减一次，则将商的对应位加1}}
OutputResult:for (int i = 0; i < MAXL; i++) {if (Result[i] >= 10) {Result[i + 1] += Result[i] / 10;Result[i] %= 10;}}bool bStartOutput = false;for (int i = MAXL; i >= 0; i--) {if (bStartOutput)printf("%d", Result[i]);else if (Result[i]) {printf("%d", Result[i]);bStartOutput = true;}}if (!bStartOutput)printf("0");return 0;
}
7.4 麦森数
我输了，我还是不会做
主要的思想是用到了快速幂
#include <cstdio>
#include <math.h>
#include <memory.h>
#define LEN 125/*
Multiply函数功能是计算高精度乘法a*b
结果的末500未放在a中
*/
void Multiply(int* a, int* b) {int nCarry;//进位int nTmp;int c[LEN];//存放结果的末500位memset(c, 0, sizeof(int)*LEN);for (int i = 0; i < LEN; i++) {nCarry = 0;for (int j = 0; j < LEN - i; j++) {nTmp = c[i + j] + a[j] * b[i] + nCarry;c[i + j] = nTmp % 10000;nCarry = nTmp / 10000;}}memcpy(a, c, LEN * sizeof(int));
}int main() {int p;int anPow[LEN];//存放不断增长的2的次幂int aResult[LEN];//存放最终结果的末500位scanf("%d", &p);printf("%d\n", (int)(p*log10(2)) + 1);//将2的次幂初始化为2^(2^0)//最终结果初始化为1anPow[0] = 2;aResult[0] = 1;for (int i = 1; i < LEN; i++) {anPow[i] = 0;aResult[i] = 0;}//计算2的p次方while (p > 0) {if (p & 1)Multiply(aResult, anPow);p >>= 1;Multiply(anPow, anPow);}aResult[0]--;//2的p次方算出后减1for (int i = LEN - 1; i >= 0; i--) {if (i % 25 == 12)printf("%02d\n%02d", aResult[i] / 100, aResult[i] % 100);else {printf("%04d", aResult[i]);if (i % 25 == 0)printf("\n");}}return 0;
}
8.2 生理周期
想太复杂了。。。以为在给定的day前面也可能会有，然后输出一个负的，结果一直WA……直接从day+1开始判断，AC……
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;int main() {int tili, ganqing, zhili;scanf("%d %d %d", &tili, &ganqing, &zhili);int day;scanf("%d", &day);for (int i = day+1; i <= 21252+day; i++) {if (abs(i - tili) % 23 == 0 && abs(i - ganqing) % 28 == 0 && abs(i - zhili) % 33 == 0) {printf("%d", i - day);break;}}return 0;
}
8.3 称硬币
写了大半天，WA一百遍，然后发现自己判断硬币的逻辑出了问题。
把书里的正确的代码贴上来，也是复习了。
另，把自己错误的代码放在后面，分析错误所在。
#include <cstdio>
#include <string.h>char left[3][7], right[3][7], result[3][5];bool isLight(char x) {for (int i = 0; i < 3; i++) {switch (result[i][0]) {case'u':if (strchr(right[i], x) == NULL)return false;break;case'e':if (strchr(right[i], x) != NULL || strchr(left[i], x) != NULL)return false;break;case'd':if (strchr(left[i], x) == NULL)return false;break;}}return true;
}bool isHeavy(char x) {for (int i = 0; i < 3; i++) {switch (result[i][0]) {case'u':if (strchr(left[i], x) == NULL)return false;break;case'e':if (strchr(right[i], x) != NULL || strchr(left[i], x) != NULL)return false;break;case'd':if (strchr(right[i], x) == NULL)return false;break;}}return true;
}int main() {int n;char c;scanf("%d", &n);while (n--) {for (int i = 0; i < 3; i++)scanf("%s %s %s", left[i], right[i], result[i]);for (c = 'A'; c <= 'L'; c++) {if (isLight(c)) {printf("%c is the counterfeit coin and it is light.\n", c);break;}if (isHeavy(c)) {printf("%c is the counterfeit coin and it is heavy.\n", c);break;}}}return 0;
}
错误代码：
错误举例：这种判断方法，假如j是假币，前两次i和j一起称，最后一次只称了j。这样会判断i是假币。
#include <iostream>
#include <string>
using namespace std;int main() {int n;scanf("%d", &n);while (n--) {string left[3], right[3], result[3];for (int i = 0; i < 3; i++)cin >> left[i] >> right[i] >> result[i];for (char temp='A'; temp <= 'L'; temp++) {int Isfalse = -1;//0表示更轻，1表示更重，2表示是真币bool flag = false;//有没有出现for (int j = 0; j < 3; j++) {if (Isfalse == 2)break;if (left[j].find(temp) != string::npos || right[j].find(temp) != string::npos) {flag = true;if ((left[j].find(temp) != string::npos && result[j][0] == 'e') || (right[j].find(temp) != string::npos && result[j][0] == 'e'))Isfalse = 2;if ((left[j].find(temp) != string::npos && result[j][0] == 'u') || (right[j].find(temp) != string::npos && result[j][0] == 'd'))Isfalse = 1;if ((left[j].find(temp) != string::npos && result[j][0] == 'd') || (right[j].find(temp) != string::npos && result[j][0] == 'u'))Isfalse = 0;}}if (flag&&Isfalse != 2) {if (Isfalse == 0)printf("%c is the counterfeit coin and it is light.\n", temp);else if (Isfalse == 1)printf("%c is the counterfeit coin and it is heavy.\n", temp);break;}}}return 0;
}
8.4 完美立方
#include <cstdio>int main() {int N;scanf("%d", &N);int a, b, c, d;for (a = 2; a <= N; a++) {for (b = 2; b <= N; b++) {for (c = b; c <= N; c++) {for (d = c; d <= N; d++) {if (a*a*a < b*b*b + c * c*c + d * d*d)break;if (a*a*a > b*b*b + c * c*c + d * d*d)continue;if (a*a*a == b * b*b + c * c*c + d * d*d)printf("Cube = %d, Triple = (%d,%d,%d)\n", a, b, c, d);}}}}return 0;
}
8.5 熄灯问题
#include <cstdio>
int puzzle[6][8] = { 0 }, press[6][8] = { 0 };bool guess() {for (int r = 1; r < 5; r++) {for (int c = 1; c < 7; c++)press[r + 1][c] = (puzzle[r][c] + press[r][c] + press[r - 1][c] + press[r][c - 1] + press[r][c + 1]) % 2;}for (int c = 1; c < 7; c++)if (puzzle[5][c] != (press[5][c] + press[5][c - 1] + press[5][c + 1] + press[4][c]) % 2)return false;return true;
}void enumate() {int c;for (c = 1; c < 7; c++)press[1][c] = 0;while (guess() == false) {press[1][1]++;c = 1;while (press[1][c] > 1) {press[1][c] = 0;c++;press[1][c]++;}}return;
}int main() {for (int i = 1; i <= 5; i++)for (int j = 1; j <= 6; j++)scanf("%d", &puzzle[i][j]);enumate();for (int i = 1; i <= 5; i++) {for (int j = 1; j <= 6; j++)printf("%d ", press[i][j]);printf("\n");}
}
8.6 恼人的小青蛙
#include <cstdio>
#include <stdlib.h>
#include <algorithm>>
using namespace std;struct PLANT {int x, y;
}plants[5010];int row, col;
int n;bool cmp(PLANT a, PLANT b) {if (a.x != b.x)return a.x < b.x;elsereturn a.y < b.y;
}int SearchPath(PLANT secPlant, int dx, int dy) {PLANT plant;int steps;plant.x = secPlant.x + dx;plant.y = secPlant.y + dy;steps = 2;while (plant.x <= row && plant.x >= 1 && plant.y <= col && plant.y >= 1) {if (!binary_search(plants, plants + n, plant, cmp)) {steps = 0;break;}plant.x += dx;plant.y += dy;steps++;}return steps;
}int main() {scanf("%d %d", &row, &col);int dx, dy, px, py, steps = 0, max = 2;scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d %d", &plants[i].x, &plants[i].y);sort(plants, plants + n,cmp);for (int i = 0; i < n - 2; i++) {for (int j = i + 1; j < n - 1; j++) {dx = plants[j].x - plants[i].x;dy = plants[j].y - plants[i].y;px = plants[i].x - dx;py = plants[i].y - dy;if (px <= row && px >= 1 && py <= col && py >= 1)continue;if (plants[i].x + (max - 1)*dx > row)break;py = plants[i].y + (max - 1)*dy;if (py > col || py < 1)continue;steps = SearchPath(plants[j], dx, dy);if (steps > max)max = steps;}}if (max == 2)max = 0;printf("%d", max);return 0;
}
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