2020牛客国庆集训派对day2（A，C，D，E，F，G，H，J，）
2024-04-25 16:06:30
A、AKU NEGARAKU
约瑟夫环裸题
AC代码
#include<iostream>
#include<stdio.h>
using namespace std;
#define ll long long
ll yuesefu(ll n, ll m) {if (n == 1) {return 0; }else {return (yuesefu(n - 1, m) + m) % n;}
}
int main(void) {int a, b;while (cin >> a >> b && a + b != 0) {cout << yuesefu(a, b) + 1 << endl;}return 0;
}C、ELI’S CURIOUS MIND
一个假的dp
AC代码
#include<stdio.h>
#define ll long long
int main()
{int n, t = 1;ll dp[100];dp[0] = 0; dp[1] = 1; dp[2] = 1;for (int i = 3; i < 80; i++)dp[i] = dp[i - 2] + dp[i - 3];dp[1] = 0; dp[2] = 0;while (scanf("%d", &n) && n) {printf("Case #%d: ", t++);ll res = dp[n] + dp[n - 1];printf("%lld\n", res);}return 0;
}
D、EXPLORACE
最小生成树裸题
AC代码
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n, m, tot = 0, k = 0;
int fat[2000];
struct node
{int from, to, dis;
}edge[200];
bool cmp(const node& a, const node& b)
{return a.dis < b.dis;
}
int father(int x)
{if (fat[x] != x)return father(fat[x]);else return x;
}
void unionn(int x, int y)
{fat[father(y)] = father(x);
}
int main()
{int t; scanf("%d", &t);int nn = 1;while (t--) {//memset(fat, 0, sizeof(fat));scanf("%d%d", &n, &m); for (int i = 1; i <= m; i++){scanf("%d%d%d", &edge[i].from, &edge[i].to, &edge[i].dis);}k = 0, tot = 0;for (int i = 1; i <= n; i++) fat[i] = i;sort(edge + 1, edge + 1 + m, cmp);//按权值排序（kruskal的体现） for (int i = 1; i <= m; i++)//从小到大遍历 {if (k == n - 1) break;//n个点需要n-1条边连接 if (father(edge[i].from) != father(edge[i].to)){unionn(edge[i].from, edge[i].to);//加入 tot += edge[i].dis;//记录边权 k++; }}printf("Case #%d: %d meters\n", nn++, tot);}return 0;
}
E、MATRIX MULTIPLICATION CALCULATOR
矩阵相乘，暴力就可以过
AC代码
#include<stdio.h>
#include<string.h>
int a[25][25], b[25][25], c[25][25];
int main()
{int x, y, x1, y1;int t = 1;while (scanf("%d%d%d%d", &x, &y, &x1, &y1) && x + y + x1 + y1 != 0) {for (int i = 0; i < x; i++) {for (int j = 0; j < y; j++)scanf("%d", &a[i][j]);}for (int i = 0; i < x1; i++) {for (int j = 0; j < y1; j++)scanf("%d", &b[i][j]);}printf("Case #%d:\n", t++);if (y != x1) {printf("undefined\n");continue;}memset(c, 0, sizeof(c));for (int i = 0; i < x; i++) {for (int j = 0; j < y1; j++) {for (int k = 0; k < y; k++) {c[i][j] += a[i][k] * b[k][j];}}}for (int i = 0; i < x; i++) {printf("| ");for (int j = 0; j < y1; j++) {printf("%d ", c[i][j]);}printf("|\n");}}return 0;
}
F、SUM OF SUB RECTANGLE AREAS
推公式，由于是大数，拿python写的
AC代码
t = int(input())
for i in range(t):n = int(input())res = n * n - (n - 1) * n // 2 + n ** 3 - n * n * (n - 1) + (n - 1) * n * (2 * n - 1) // 6res //= 2res *= resprint(res)
G、WAK SANI SATAY
队友写的(非本人代码，仅供参考)，买菜题
AC代码
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<string>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<iomanip>
using namespace std;#define getT long long t;scanf("%lld",&t)
#define getN long long n;scanf("%lld",&n)
#define for0n(n) getN; for(long long i=0;i<n;i++)
#define for1n(n) getN; for(long long i=1;i<=n;i++)
#define for0(n) for(long long i=0;i<n;i++)
#define for1(n) for(long long i=1;i<=n;i++)
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)typedef long long ll;
ll mod;
ll gcd(ll a, ll b) {return a % b == 0 ? b : gcd(b, a%b);
}ll pow(ll a, ll b) {ll an = 1;ll base = a % mod;while (b > 0) {if (b & 1 != 0) {an = (an*base) % mod;}base = (base*base) % mod;b >>= 1;}return an % mod;
}ll inv(ll a) {return pow(a, mod - 2);
}
ll lcm(ll a, ll b) {return a * b / gcd(a, b);
}
ll Arithmetic(ll a1, ll n, ll d) {ll sum = 0;sum += n * a1;sum %= mod;ll temp = n * (n - 1);temp %= mod;temp *= d;temp %= mod;temp /= 2;temp %= mod;return (sum + temp) % mod;
}
ll euler(ll n)
{ll ans = n;for (ll i = 2; i*i <= n; i++)if (n%i == 0){ans = ans - ans / i;while (n%i == 0)n /= i;}if (n > 1)ans = ans - ans / n;return ans;
}
int main(void) {getN;ll cnt = 1;const double base_num_stick = 85.0;const double rate_a = 0.8-(15.5 / base_num_stick), rate_b =1.0-( 32.0 / base_num_stick), rate_c = 1.20-(40.0 / base_num_stick);while (n){double a, b, c, d;double total = 0.0;double ans = 0;for0(n) {scanf("%lf%lf%lf%lf", &a, &b, &c, &d);total += a;total += b;total += c;ans += a * rate_a;ans += b * rate_b;ans += c * rate_c;ans += 0.60*d;}//ans -= total / 85.0*8.0;printf("Case #%lld: RM%.2lf\n", cnt++, ans);scanf("%lld", &n);}return 0;
}
H、STROOP EFFECT
将所有情况按照题目存在一个表中，只要现在这个新表和题目中的表成比例，即可刺激
AC代码
#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
int main()
{int T; scanf("%d", &T);int tbb[5][5];for (int ca = 1; ca <= T; ca++) {int x, m = 0, n = 0, nm = 0, nn = 0;int tb[5][5];memset(tb, 0, sizeof(tb));int flag = 0;while (scanf("%d", &x) && x) {int a = x / 10;int b = x % 10;tb[a][b]++;if (a != m) {nm = 1;m = a;}elsenm++;if (b != n) {nn = 1;n = b;}elsenn++;if (nm >= 3 || nn >= 3)flag = 1;}if (flag) {printf("Case #%d: Not Stroop\n", ca);continue;}int p;if (tb[1][1] % 3 == 0) {p = tb[1][1] / 3;}else {printf("Case #%d: Not Stroop\n", ca);continue;}int f = 0;for (int i = 1; i <= 4; i++) {for (int j = 1; j <= 4; j++) {if (i == j) {if (tb[i][j] % 3 == 0) {if (p != tb[i][j] / 3)f = 1;}elsef = 1;}else {if (p != tb[i][j])f = 1;}}}if (f)printf("Case #%d: Not Stroop\n", ca);elseprintf("Case #%d: Stroop\n", ca);}return 0;
}
J、VIRUS OUTBREAK
大数斐波那契数列
AC代码
import java.math.BigInteger;
import java.util.Scanner;
public class Main{public static void main(String args[]){BigInteger a[] = new BigInteger[500];Scanner cin = new Scanner(System.in);a[1] = BigInteger.valueOf(1);a[2] = BigInteger.valueOf(1);for(int i = 3; i < 495; i++)a[i] = a[i - 1].add(a[i - 2]);int n = cin.nextInt();while(n != -1){System.out.println("Hour: " + n + ": " + a[n] + " cow(s) affected");n = cin.nextInt();}}
}2020牛客国庆集训派对day2（A，C，D，E，F，G，H，J，）相关推荐
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