uva10487-最接近的和

题目链接 http://acm.hust.edu.cn/vjudge/problem/21350

解题思路

先排序。

对于每个数，找到另外两个数，使得与其中一个加起来小于目标与另外一个加起来大于目标。

如果找不到可能是只能选头两个或末尾两个。

再求绝对值那个更小就行了。

代码

#include<stdio.h>
#include<algorithm>
#define MAX_NUM 1<<31-1
using namespace std;
const int maxLen = 1005;
int data[maxLen];
int targer[26];
int ans[26];
int minv;
int Judge(int i, int n, int t)
{
    int pre = -1;
    for(int k=i+1; k<n-1; k++)
        if(data[k]+data[i]<t && data[k+1]+data[i]>=t) pre=k;
    if(pre==-1) {
        if(data[i]+data[n-1]<t) {
            if(abs(data[i]+data[n-1]-t)<minv) {
                   minv = abs(data[i]+data[n-1]-t);
                return data[i]+data[n-1];
            }
            else return MAX_NUM;
        }
        else if(data[i]+data[i+1]>=t) {
            if(abs(data[i]+data[i+1]-t)<minv) {
                minv = abs(data[i]+data[i+1]-t);
                return data[i]+data[i+1];
            }
            else return MAX_NUM;
        }
    }
    else if(abs(data[i]+data[pre]-t)<abs(data[i]+data[pre+1]-t)) {
            if(minv>abs(data[i]+data[pre]-t))
                { minv = abs(data[i]+data[pre]-t); return data[i]+data[pre]; }
            else return MAX_NUM;
        }
        else
            if(minv>abs(data[i]+data[pre+1]-t))
                { minv = abs(data[i]+data[pre+1]-t); return data[i]+data[pre+1]; }
            else return MAX_NUM;
}
int main()
{
    int n, cases = 1;
    scanf("%d", &n);
    while(n != 0) {
        for(int i=0; i<n; i++) scanf("%d", &data[i]);
        sort(data, data+n);
        int m;
        scanf("%d", &m);
        for(int i=0; i<m; i++) {
            scanf("%d", &targer[i]);
            minv = MAX_NUM;
            for(int j=0; j<n-1; j++) {
                int a = Judge(j,n,targer[i]);
                if(a != MAX_NUM)
                    ans[i] = a;
            }
        }
        printf("Case %d:\n", cases);
        for(int i=0; i<m; i++)
            printf("Closest sum to %d is %d.\n", targer[i], ans[i]);
        cases++; scanf("%d", &n);
    }
    return 0;
}