2021蓝桥杯预选赛题解

链接：https://witacm.com/problemset.php?p=11

A : 过了几天

写时间或者日期内的题目时先把单位统一化成低级单位，计算完成后再化成标准单位。

# include <iostream>struct date {int year;int month;int day;
};int mon[20] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};bool isLeap(int year) {return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}int cnt(date a) {int y = a.year;int m = a.month;int d = a.day;int sum = 0;for (int i = 1; i < y; i++) {if (isLeap(i)) {sum += 366;} else {sum += 365;}}for (int i = 1; i < m; i++) {if (i == 2) {if (isLeap(y)) {sum += 29;} else {sum += 28;}} else {sum += mon[i];}}sum += d - 1;return sum;
}void solve() {date d1, d2;char c;std::cin >> d1.year >> c >> d1.month >> c >> d1.day>> d2.year >> c >> d2.month >> c >> d2.day;int num1 = cnt(d1);int num2 = cnt(d2);std::cout << num2 - num1 + 1 << "\n";
}int main() {int T;std::cin >> T;while (T--) {solve();}return 0;
}

B. 排列数字

# include <iostream>
# include <algorithm>bool cmp(int a, int b) {return a > b;
}int main() {int a[10] = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1};int n, k;std::cin >> n >> k;int cnt = 0;do {cnt++;if (cnt == k) {break;}} while (std::next_permutation(a + 10 - n, a + 10, cmp));for (int i = 10 - n; i < 10; i++) {std::cout << a[i] << " ";}return 0;
}

# include <iostream>
# include <algorithm>int main() {int a[10] = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1};int n, k;std::cin >> n >> k;int cnt = 0;do {cnt++;if (cnt == k) {break;}} while (std::next_permutation(a + 10 - n, a + 10, std::greater<int>()));for (int i = 10 - n; i < 10; i++) {std::cout << a[i] << " ";}return 0;
}

# include <iostream>
# include <algorithm>struct d{int num;bool operator < (d rhs) {return this->num > rhs.num;}
};int main() {d a[10];for (int i = 0; i < 10; i++) {a[i].num = 10 - i;}int n, k;std::cin >> n >> k;int cnt = 0;do {cnt++;if (cnt == k) {break;}} while (std::next_permutation(a + 10 - n, a + 10));for (int i = 10 - n; i < 10; i++) {std::cout << a[i].num << " ";}return 0;
}

#include <iostream>
#include <algorithm>int a[15] = {0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
int choose[15];
bool vis[15];int n, k;
int cnt = 0;void dfs(int step) {if (cnt > k) {return;} if (step == n + 1) {cnt++;if (cnt == k) {for (int i = 1; i <= n; i++) {std::cout << choose[i] << " ";}return;}return;}for (int i = n; i >= 1; i--) {if (!vis[i]) {vis[i] = true;choose[step] = a[10 - i + 1];dfs(step + 1);vis[i] = false;}}
}int main() {std::cin >> n >> k;dfs(1);return 0;
}

C. Simple in Naive

三关键字排序
1.

# include <iostream>
# include <set>
# include <algorithm>
# include <cstdio>struct person {int id;char c;int num;bool operator < (person rhs) {if (c != rhs.c) {return c < rhs.c;} else {if (num != rhs.num) {return num > rhs.num;               }return id < rhs.id;}}
};const int maxn = 1e5 + 5;
person p[maxn];
std::set<int> st;void solve1() {int Q;scanf("%d", &Q);int t;while (Q--) {scanf("%d", &t);puts("YES");}
}int main()  {int n, T;std::cin >> n >> T;for (int i = 1; i <= n; i++) {scanf(" %c%d", &p[i].c, &p[i].num);p[i].id = i;}if (T >= n) {solve1();return 0;}std::sort(p +  1, p + n + 1);for (int i = 1; i <= T; i++) {st.insert(p[i].id);}int Q;std::cin >> Q;while (Q--) {int t;scanf("%d", &t);if (st.find(t) != st.end()) {puts("YES");} else {puts("NO");}}return 0;
}

# include <iostream>
# include <algorithm>
# include <cstdio>struct person {int id;char c;int num;bool operator < (person rhs) {if (c != rhs.c) {return c < rhs.c;} else {if (num != rhs.num) {return num > rhs.num;               }return id < rhs.id;}}
};const int maxn = 1e5 + 5;
person p[maxn];
bool vis[maxn];void solve1() {int Q;scanf("%d", &Q);int t;while (Q--) {scanf("%d", &t);puts("YES");}
}int main()  {int n, T;std::cin >> n >> T;for (int i = 1; i <= n; i++) {scanf(" %c%d", &p[i].c, &p[i].num);p[i].id = i;}if (T >= n) {solve1();return 0;}std::sort(p +  1, p + n + 1);for (int i = 1; i <= T; i++) {vis[p[i].id] = true;}int Q;std::cin >> Q;while (Q--) {int t;scanf("%d", &t);if (vis[t]) {puts("YES");} else {puts("NO");}}return 0;
}

D. 千禧之数

改编自武汉工程大学计算机科学与工程学院第二届程序设计新生赛（预选赛）的D题

#include <iostream>typedef long long ll;int main()
{int n;std::cin >> n;ll sum = 0;for (int j = 2; j <= n / j; j++){if (n % j == 0){sum += j;while (n % j == 0){n /= j;}}}if (n > 1) {sum += n;}if (sum >= 2000){std::cout << "Yes";}else{std::cout << "No";}return 0;
}

E. 喜新厌旧

改编自武汉工程大学计算机科学与工程学院第二届程序设计新生赛的J题

#include <iostream>
#include <algorithm>
#include <string>
#include <cstdio>std::string t;
std::string s;int main()
{std::cin >> t >> s;std::sort(t.begin(), t.end());int j = 0;for (int i = 0; i < s.size(); i++){if (s[i] <= t[j]){putchar(s[i]);continue;}else{putchar(t[j++]);}}return 0;
}

F. 探亲

改编自武汉工程大学第一届程序设计女生赛的C题
邀请码 2020wfinal
两个bfs

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;typedef pair<int,int> PII; const int N=1010;
int  m,n;
queue<PII> q;
char g[N][N];
int d[N][N];int bfs(int x1,int y1,int x2,int y2){memset(d,-1,sizeof d);while (q.size()) {q.pop();}q.push({x1,y1});d[x1][y1]=0;while(q.size()){PII t=q.front();q.pop();if(t.first==x2 && t.second==y2) return d[x2][y2];int dx[]={-1,0,1,0};int dy[]={0,-1,0,1};for(int i=0;i<4;i++){int x=t.first+dx[i];int y=t.second+dy[i];if(x>0 && x<=n && y>0 && y<=m && g[x][y]=='.' && d[x][y]==-1){d[x][y]=d[t.first][t.second]+1;q.push({x,y});}}}return d[x2][y2];
}int main(){int T;cin>>T;while(T--){cin>>n>>m;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)cin>>g[i][j];int x1,y1,x2,y2,x3,y3;cin>>x1>>y1>>x2>>y2>>x3>>y3;int res1=bfs(x1,y1,x2,y2);int res2=bfs(x2,y2,x3,y3);//cout<<res1<<" "<<res2<<" ";if(res1==-1 || res2== -1) cout<<"-1"<<endl;else cout<<res1+res2<<endl;}return 0;
}

G. 计算器

搬运自2019年武汉科技大学菜鸟杯
模拟即可

#include<stdio.h>
#include<string.h>
const int MAXN=1e6+5;
char s[MAXN];
char ss[MAXN];
int a,n;
int main()
{scanf("%s%d",s+1,&n);int len=strlen(s+1);for(int i=1;i<=len;i++)a=a*10+s[i]-'0';for(int i=1;i<=n;i++){scanf("%s",ss+1);for(int j=1;j<=9;j++){int flag=1;for(int k=1;k<=len;k++){int x=s[k]-'0',y=ss[k]-'0';if(x*j%10!=y){flag=0;break;}}if(flag){a*=j;break;}}strcpy(s+1,ss+1);}printf("%d",a);
}

H. ZZZ的银河女装传说

搬运自2019年武汉科技大学菜鸟杯

题意可以转化成求最少变换次数让字符串变为左边一串连续的’.‘序列，右边一串连续的’#‘序列，或者全变为’.‘序列或者’#‘序列。
我们可以预处理出’.‘和’#‘的前缀和,a[i]代表前i个字符里有a[i]个’.’, b[i]代表前i个字符里有b[i]个’#’，那么可以直接遍历字符串，枚举每个点为’.‘和’#‘序列的分界点(’.‘的最后一个点)时的情况。要想让衣柜是整齐的，那么只需要把分界点前的’#‘全部变为’.’ (贡献为b[i]),将分界点后的’.‘全部变为’#’ (贡献为a[n]-a[i])。每一次枚举计算答案是O(1)的，所以总复杂度为O(n)。
ans=min(ans,b[i]+a[n]−a[i])ans=min(ans,b[i]+a[n]-a[i])ans=min(ans,b[i]+a[n]−a[i])

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define rg register
#define ll long long
#define CL(a,b) memset(a,b,sizeof a)
const int N=2e5+7,inf=0x3f3f3f3f;
char s[N];
int a[N];
int black[N],white[N];
int main()
{int n;scanf("%d",&n);scanf("%s",s);for(int i=0;i<n;i++)if(s[i]=='.')a[i+1]=0;else a[i+1]=1;int ans=inf;for(int i=1;i<=n;i++)if(a[i]==0){white[i]=white[i-1]+1;black[i]=black[i-1];}else{black[i]=black[i-1]+1;white[i]=white[i-1];}for(int i=0;i<=n;i++)ans=min(black[i]+white[n]-white[i],ans);cout<<ans<<endl;return 0;
}

论往届题目的重要性