10125-Sumsets【暴力】

利用n^2的时间枚举所有a[i] + a[j]

利用n^2的时间枚举所有a[i] - a[j]

之后利用n^2时间一个一个找a[i] - a[j]的值是否存在于a[i] + a[j]中

找的时候需要二分查找

另外一点就是注意long long的范围以及四个数是集合内不同的四个元素

15222638

10125

Sumsets

Accepted

C++

0.449

2015-03-26 15:40:29

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1005;
struct St{
    LL v;
    int x,y;
    St(LL v,int x,int y):v(v),x(x),y(y){};
};
int n,ok;
LL ans;
LL array[maxn];
vector<St>arr;
vector<St>arr2;
void debug(){
    for(int i = 0; i < arr.size(); i++)
        printf("%I64d ",arr[i].v);
    puts("");
    for(int i = 0; i < arr2.size(); i++)
        printf("%I64d ",arr2[i].v);
    puts("");
}
bool cmp1(St a,St b){
    return a.v < b.v;
}
bool cmp2(St a,St b){
    return a.v > b.v;
}
bool solve(LL v,int x,int y){
    int l = 0, r = arr.size() - 1;
    while(l < r){
        int mid = l + (r - l) / 2;
        //printf("%I64d %d %d %d %I64d\n",arr[mid].v,l,r,mid,v);
        if(arr[mid].v >= v)
            r = mid;
        else
            l = mid + 1;
    }
    for(int i = l; i < arr.size(); i++){
        if(arr[i].v != v) break;
        int  xx = arr[i].x,yy = arr[i].y;
        if(xx != x && xx != y && yy != x && yy != y){
            if(!ok)
                ans = arr[i].v + array[y];
            else
                ans = max(ans,arr[i].v + array[y]);
            return true;
        }
    }
    return false;
}
int main(){
    while(scanf("%d",&n) && n){
        arr.clear();
        arr2.clear();
        for(int i = 0; i < n; i++)
            scanf("%lld",&array[i]);
        for(int i = 0; i < n; i++)
            for(int j = i + 1; j < n; j++)
                    arr.push_back(St(array[i] + array[j],i,j));
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                if(i != j)
                    arr2.push_back(St(array[i] - array[j],i,j));
        sort(arr.begin(),arr.end(),cmp1);
        sort(arr2.begin(),arr2.end(),cmp2);
        //debug();
        ok = 0;
        for(int i = 0; i < arr2.size(); i++){
            LL v = arr2[i].v;
            int x = arr2[i].x,y = arr2[i].y;
            if(solve(v,x,y)){
                ok = 1;
            }
        }
        if(ok)
            printf("%lld\n",ans);
        else
            printf("no solution\n");
    }
    return 0;
}