标签：size str 思路 style 排列 输入 第一天 must pac
There are n kinds of goods in the company, with each of them has a inventory of cnti and direct unit benefit vali. Now you find due to price changes, for any goods sold on day i, if its direct benefit is val, the total benefit would be i?val.
Beginning from the first day, you can and must sell only one good per day until
you can‘t or don‘t want to do so. If you are allowed to leave some goods
unsold, what‘s the max total benefit you can get in the end?
The first line contains an integers n(1≤n≤1000).
The second line contains n integers val1,val2,..,valn(?100≤vali.≤100).
The third line contains n integers cnt1,cnt2,..,cntn(1≤cnti≤100).
Output an integer in a single line, indicating the max total benefit.
Hint: sell goods whose price with order as -1, 5, 6, 6, the total benefit would be -1*1 + 5*2 + 6*3 + 6*4 = 51.
4 -1 -100 5 6 1 1 1 2
51
思路：把商品价值从小到大排列 然后从后向前扫 sum 记录已经扫过的商品的总价值（val） ans 记录已经卖出的 商品价值（i*val），
#include <cstdio> #include <iostream> #include <algorithm> using namespace std ; #define maxn 1050 #define LL long long struct node { int val ; int cnt ; }; node num[maxn] ; bool cmp(node a , node b ){ return a.val < b.val ; } int main(){ int n ; LL ans , sum ; while(~scanf("%d" , &n)){ sum = ans = 0 ; for(int i=0 ; i<n ; i++){ scanf("%d" , &num[i].val) ; } for(int i=0 ; i<n ; i++){ scanf("%d" , &num[i].cnt) ; } sort(num , num +n , cmp) ; for(int i=n-1 ; i>=0 ; i--){ for(int j=0 ; j<num[i].cnt ; j++){ if(ans + sum + num[i].val >=ans){ ans = ans + sum + num[i].val ; sum = sum + num[i].val ; } } } printf("%lld\n" , ans) ; //注意 ans 必须开 LL } return 0 ; }
标签：size str 思路 style 排列 输入 第一天 must pac
原文地址：http://www.cnblogs.com/yi-ye-zhi-qiu/p/7634469.html
