UVA 10125 (14.3.6)

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Problem C - Sumsets

Given S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S.

Input

Several S, each consisting of a line containing an integer 1 <= n <= 1000indicating the number of elements in S, followed by the elements of S, one per line. Each element of S is a distinct integer between -536870912 and +536870911 inclusive.The last line of input contains 0.

Output

For each S, a single line containing d, or a single line containing "no solution".

Sample Input

52 3 5 7 1252 16 64 256 10240

Output for Sample Input

12no solution

题意：S个数，从中选出四个，要符合a+b+c = d

思路：

首先，如果暴力枚举，那么有四层循环，看看数据大小，必超时～

那我们换思路，大体上，S个数先存在数组排序排一下，再利用四个数的相互制约的关系来解题，减少不必要的尝试

排序简单，sort一下即可，制约关系就要找了，之前我们排序过，那第一个数是最小的，不可能是d吧?

So，我们明白了d是和，大于a b c中的任何一个数，那么我们就从数组中的最后一个开始枚举好了

接下来，我们假设a < b < c 那么我们得出 d - c = a + b对不?

这样就清晰了, 我们从数组后端从大到小枚举c 和 d, 并满足c < d; 然后从数组前端从小到大枚举a 和 b, 并满足a < b

剩下的还有一步对于a和b的特殊处理, 交给你们自己代码意会~

AC代码:

#include<stdio.h>#include<algorithm>using namespace std;int num[1005];int main() {    int s;    while(scanf("%d", &s) != EOF) {        if(s == 0)            break;        for(int i = 0; i < s; i++)            scanf("%d", &num[i]);        sort(num, num+s);                int a, b, c, d;        int judge = 0;        for(d = s-1; d >= 0; d--) {            int ans;            for(c = s-1; c > 1; c--) {                if(c != d)                    ans = num[d] - num[c];                for(a = 0, b = c-1; a < b;) {                    if(num[a] + num[b] == ans) {                        judge = 1;                        break;                    }                    else if(num[a] + num[b] < ans)                        a++;                    else                        b--;                }                if(judge)                    break;            }            if(judge)                break;        }        if(judge)            printf("%d\n", num[d]);        else            printf("no solution\n");    }    return 0;}

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