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  • CodeForces 722A Broken Clock (水题)

    题意:给定一个时间,然后改最少的数字,使得它成为12进制或者24进制。

    析:24进制主要判是不是大于23,如果是把第一位变成0,12进制判是不是大于12,如果是再看第二位是不是0,是0,第一位变成1,不是第一位变成0,

    如果等于0,第一位变成1,至于分钟,只要判是不是大于59,是把第一位变成0.

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    //#include <tr1/unordered_map>
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    //using namespace std :: tr1;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const LL LNF = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e5 + 5;
    const LL mod = 1e9 + 7;
    const int N = 1e6 + 5;
    const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
    const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
    const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
    inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline int Min(int a, int b){ return a < b ? a : b; }
    inline int Max(int a, int b){ return a > b ? a : b; }
    inline LL Min(LL a, LL b){ return a < b ? a : b; }
    inline LL Max(LL a, LL b){ return a > b ? a : b; }
    inline bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    
    int main(){
        while(scanf("%d", &n) == 1){
            string s;
            cin >> s;
            int h = (s[0] - '0') * 10 + s[1] - '0';
            int m = s[3] - '0';
            if(m > 5)  s[3] = '0';
            if(n == 24){
                if(h > 23)  s[0] = '0';
            }
            else{
                if(h > 12){
                    if(s[1] != '0')  s[0] = '0';
                    else s[0] = '1';
                }
                if(!h) s[0] = '1';
            }
            cout << s << endl;
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5930133.html
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