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  • C. Counting Pair

    题目链接:戳这里

    题意:男生编号1~N,女生编号1~M,给一个sum,问有多少对男女编号加起来等于sum
    思路:若sum≤min(n,m),那么有sum-1对(1&sum-1、2&sum-2……);
    若sum>max(n,m),那么有N+M-sum+1对(N&sum-N、N-1&sum-N+1……sum-M&M);
    否则有min(n,m)对(设n≤m,1&N-1、2&n-2……)

    AC Code:

     1 #include <cstdio>
     2 #include <algorithm>
     3 using namespace std;
     4 
     5 int main() {
     6     int T, n, m, Q, x, ans;
     7     scanf("%d", &T);
     8     for(int t = 1; t <= T; ++t) {
     9         scanf("%d%d%d", &n, &m, &Q);
    10         printf("Case #%d:
    ", t);
    11         for(int i = 0; i < Q; ++i) {
    12             scanf("%d", &x);
    13             if(x <= n && x <= m) {
    14                 ans = x - 1;
    15             }
    16             else if(x > n && x > m) {
    17                 ans = n + m - x + 1;
    18                 if(ans < 0) ans = 0;
    19             }
    20             else {
    21                 ans = min(n, m);
    22             }
    23             printf("%d
    ", ans);
    24         }
    25     }
    26 }

    By 区彦开

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  • 原文地址:https://www.cnblogs.com/scnuacm/p/3221280.html
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