ZOJ3607-Lazier Salesgirl (暴力)

ZOJ3607-Lazier Salesgirl

题目
题意：在最短等待时间（大于每个间隔时间）内求最大卖去的商品平均价格。【注意： 同一时刻必须满足该时刻所有的客人需求
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#include<cstdio>#include<cstring>using namespace std;#define inf 0xffffffint main(){    int t,n;    int price[2000],wait[2000];    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(int i=1;i<=n;i++){            scanf("%d",&price[i]);        }        double d[2000];        d[0] = wait[0] = 0;        for(int i=1;i<=n;i++){            scanf("%d",&wait[i]);            double cha =  wait[i] - wait[i-1];            d[i] = d[i-1] > cha ? d[i-1] : cha;// wait[i] max        }        wait[n+1] = inf;        double ave_max = -inf;        double d_min = -inf;        double ave,sum = 0;        for(int i=1;i<=n;i++){            sum += price[i];            ave = sum/i;            if(ave > ave_max && wait[i+1]-wait[i] > d[i] ){ //该时刻必须满足此时所有顾客                 d_min = d[i];                ave_max = ave;            }        }        printf("%.6lf %.6lf\n",d_min,ave_max);    }    return 0;}