pku 2063 investment

#include <iostream>using namespace std;int value[41], interest[41];//1 000 000 * (1.1^40) ≈ 46 000 000//46 000 000/1000 = 46 000int v[46005];int main(){int N, n;int invest, year;scanf("%d", &N);while(N--){scanf("%d%d%d", &invest, &year, &n);for(int i = 1; i <= n; ++i)scanf("%d%d", &value[i], &interest[i]);//record记录表格的方式记录下总投资额分别为j时的最大利润// v[j+value[i]] = max{v[j+value[i]], v[j]+interest[i]};memset(v, 0, sizeof(v));for(int i = 1; i <= n; ++i)for(int j = 0; j <= 46000-value[i]/1000; ++j)if(v[j] + interest[i] > v[j+value[i]/1000])v[j+value[i]/1000] = v[j] + interest[i];int sum = invest;for(int i = 1; i <= year; ++i){int max = -1;for(int j = 0; j <= sum/1000; ++j) // 寻找0~sum/1000之间最大的收入if(v[j] > max)max = v[j];sum += max; //收入增加作为下年的总投资}printf("%d/n", sum);}return 0;}/*//Learn:1. 这种题型可以用record记录表的方式处理,记录下每一种可能的情形;2. 有时占用空间太大,可以根据题意,缩小dp数组的范围(如二维变一维,为负数时加上一个base,空间线形缩小(本题))//result:Problem: 2063 User: xiaofengsheng Memory: 572K Time: 32MS Language: G++ Result: Accepted*/