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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9997 | Accepted: 2736 |
Description
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
Output
Sample Input
5 2 3 5 7 12 5 2 16 64 256 1024 0
Sample Output
12 no solution
题解：给一个序列，让找不同的a,b,c,d在集合s中，使得a+b+c=d，如果能找到输出d，否则输出no solution；
乍一看完全没思路，也许不敢动手去写，可以选从大到小排序，枚举d,c；二分a+b等于d-c即可；
extern "C++"{
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
typedef long long LL;
void SI(int &x){scanf("%d",&x);}
void SI(double &x){scanf("%lf",&x);}
void SI(char *x){scanf("%s",x);}
//void SI(LL &x){scanf("%lld",&x);}
void PI(int &x){printf("%d",x);}
void PI(double &x){printf("%lf",x);}
void PI(char *x){printf("%s",x);}
//void PI(LL &x){printf("%lld",x);}
}
const int MAXN = 1010;
int a[MAXN];
int main(){
int n;
while(scanf("%d",&n),n){
for(int i = 0;i < n;i++)SI(a[i]);
sort(a,a + n);
int ans,flot = 0;
for(int i = n - 1;i >= 0;i--){
if(flot)break;
for(int j = n - 1;j >= 0;j--){
if(flot)break;
if(i == j)continue;
int sum = a[i] - a[j],l = 0,r = j - 1;
while(l < r){
if(a[l] + a[r] == sum && i != l && i != r){
ans = a[i];
flot = 1;
break;
}
if(a[l] + a[r] > sum)
r--;
else
l++;
}
}
}
if(flot)
printf("%d\n",ans);
else
puts("no solution");
}
return 0;
}
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原文地址：http://www.cnblogs.com/handsomecui/p/5330549.html
