UVa 10487 - Closest Sums

来源：互联网发布：java中级工程师脱产班编辑：程序博客网时间：2024/05/21 15:35

算是简单题，先算出来所有俩数的和存起来，在用二分查找找出最接近的值 ~

代码如下：

#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int cmp(const void *a, const void *b){    return *(int*)a - *(int*)b;}int a[1000 + 2], b[25 + 2], s[1000000 + 10];int main(){#ifdef test    freopen("in.txt", "r", stdin);#endif    int  m, n, num = 0;    while(cin >> m && m)    {        for(int i = 0; i < m; i++)            cin >> a[i];        qsort(a, m, sizeof(a[0]), cmp);        cin >> n;        for(int i = 0; i < n; i++)            cin >> b[i];        cout <<"Case "<<++num<<":"<<endl;        int sum = -1;        for(int i = 0; i < m; i++)            for(int j = i + 1; j < m; j++)                s[++sum] = a[i] + a[j];        qsort(s, sum + 1, sizeof(s[0]), cmp);        for(int i = 0; i < n; i++)        {            int max = sum + 1, min = 0, flag = 0;            while(max > min)            {                int mid = (max + min) / 2;                if(b[i] == s[mid])                {                    flag = 1;                    break;                }                else if(b[i] > s[mid])                    min = mid + 1;                else                    max = mid;            }            if(flag)                cout<<"Closest sum to "<<b[i]<<" is "<<b[i]<<"."<<endl;            else            {                int t = s[min];                if(min + 1 <= sum)                    t = abs(s[min + 1] - b[i]) > abs(s[max]- b[i]) ? s[max] : s[min + 1];                if(min - 1 > -1)                    t = abs(s[min - 1] - b[i]) > abs(t - b[i]) ? t : s[min - 1];                cout<<"Closest sum to "<<b[i]<<" is "<<t<<"."<<endl;            }        }    }    return 0;}