Maximum Product-UVA 11059

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题目：uva1059

暴力求解

两个循环确定子序列的头和尾，从而确定了子序列的长度，乘积需要用long long int 类型

代码如下：

#include<iostream>#include<set>#include<iterator>#include<string>#include<set>#include<typeinfo>#include<queue>#include<list>#include<algorithm>#include<cstdio>#include<cctype>#include<cstring>#include<map>#include<vector>#include<cstdlib>#include<cmath>#include<stack>#include<sstream>#include<iomanip>using namespace std;vector<long long int> max1;int main(){    int n;    long long int a[200];    int kase = 0;    while(cin>>n)    {        max1.clear();        memset(a,0,sizeof(a));        for(int i = 0;i<n;i++)            cin>>a[i];        for(int i =0;i<n;i++)        {            for(int j=i;j<n;j++)                {                 long long int count1=1;                    for(int t =i;t<=j;t++)                            count1*=a[t];                max1.push_back(count1);             }        }        sort(max1.begin(),max1.end());        vector<long long int>::iterator it=max1.end();        kase++;        printf("Case #%d: The maximum product is ",kase);        if(*(it--)>0)        cout<<*(it--)<<"."<<endl;        else            cout<<0<<"."<<endl;        if(kase)            printf("\n");    }}

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