UVA10487
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题目的意思就是先给出一组数。
然后给出要计算的数的个数,
算出上面给出的一组数中,任意两个相加,得到的值中和要计算的那个数最接近的是哪个,
枚举即可;
AC代码:
#include<iostream>#include<cmath>using namespace std;int num[1000 + 5];int main () {int n;int t;int tar;int res;int T = 1; while (cin >> n && n) {for (int i = 0 ; i < n ;i++) {cin >> num[i];}cin >> t;cout << "Case " << T++ << ":\n";for (int i = 0 ; i < t ;i++) {cin >> tar;int m = abs(num[0] + num[1] - tar);res = m + tar;for (int j = 0 ; j < n ;j++) {for (int k = j + 1; k < n ;k++) {if (abs(num[j] + num[k] - tar) < m) {m = abs(num[j] + num[k] - tar);res = num[j] + num[k];}}}cout <<"Closest sum to "<< tar << " is "<< res <<"."<<endl;}}}
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