hdu1300(水题也是dp)

题意 ：

珠宝店到珍珠批发商进货 第i种价格为p[i]的珍珠需要n个 则珍珠的结算价格为∑(n+10)*p[i] 由于没种珍珠的数量结算时都要加上10 所以有时候把便宜的珍珠换为贵的结算价格反而变少了
给你一张购买清单 珍珠价格是递增的 每种珍珠都可以替换为比它贵的 求最少总花费

题解：

状态方程 dp[i] = max { dp[j] + (sum[i] - sum[j] + 10)*p[i] } （选择某段进行替换）

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#define B(x) (1<<(x))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned ui;
const int oo = 0x3f3f3f3f;
const ll OO = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define lson rt<<1
#define rson rt<<1|1
void cmax(int& a, int b){ if (b > a)a = b; }
void cmin(int& a, int b){ if (b < a)a = b; }
void cmax(ll& a, ll b){ if (b > a)a = b; }
void cmin(ll& a, ll b){ if (b < a)a = b; }
void cmax(double& a, double b){ if (a - b < eps) a = b; }
void cmin(double& a, double b){ if (b - a < eps) a = b; }
void add(int& a, int b, int mod){ a = (a + b) % mod; }
void add(ll& a, ll b, ll mod){ a = (a + b) % mod; }
const ll MOD = 1000000007;
const int maxn = 110;
int dp[maxn], sum[maxn];
int a[maxn], p[maxn];

int main(){
	//freopen("E:\\read.txt", "r", stdin);
	int T, n;
	scanf("%d", &T);
	while (T--){
		scanf("%d", &n);
		sum[0] = 0;
		for (int i = 1; i <= n; i++){
			scanf("%d %d", &a[i], &p[i]);
			sum[i] = sum[i - 1] + a[i];
		}
		dp[0] = 0;
		for (int i = 1; i <= n; i++){
			dp[i] = oo;
			for (int j = 0; j < i; j++){
				cmin(dp[i], dp[j] + (sum[i] - sum[j] + 10) * p[i]);
			}
		}
		printf("%d\n", dp
);
	}
	return 0;
}