刷个简单的DP缓缓心情 1A #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> using namespace std; #define N 10010 vector<int>q[N]; #define LL __int64 #define INF 1e…

Description Kolya has returned from a summer camp and now he's a real communication fiend. He spends all his free time on the Web chatting with his friends via ICQ. However, lately the protocol of this service was changed once again, and Kolya's clie…

1741. Communication Fiend Time limit: 1.0 second Memory limit: 64 MB Kolya has returned from a summer camp and now he's a real communication fiend. He spends all his free time on the Web chatting with his friends via ICQ. However, lately the protocol…

题目传送门 /* 题意:程序从1到n版本升级,正版+正版->正版,正版+盗版->盗版,盗版+盗版->盗版 正版+破解版->正版,盗版+破解版->盗版 DP:每种情况考虑一遍,递推就行了 注意:开long long */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using…

Description We have received an order from Pizoor Communications Inc. for a special communication system. The system consists of several devices. For each device, we are free to choose from several manufacturers. Same devices from two manufacturers d…

We have received an order from Pizoor Communications Inc. for a special communication system. The system consists of several devices. For each device, we are free to choose from several manufacturers. Same devices from two manufacturers differ in the…

个人心得:百度推荐的简单DP题,自己做了下发现真得水,看了题解发现他们的思维真得比我好太多太多, 这是一段漫长的锻炼路呀. 关于这道题,我最开始用DP的思路,找子状态,发现自己根本就不会找DP状态数组建立,怎么找都是被后面的给打乱了, 看到了网上的DP[i][j],i表示前几个设备,宽带为j的最少花费,哇,真得厉害,突然发现动态规划并不需要做到问题很完美, 其实只要解决方案能够顺带把这个解决就好了. 他的这个转移方程就是,当i等于1时,输入的想,输入的x,y(x表示宽带长度,y表示价格)dp[i…

Problem Statement      The pony Rainbow Dash wants to choose her pet. There are N animals who want to be her pet. Rainbow Dash numbered them 0 through N-1. To help her make the decision, Rainbow Dash decided to organize a relay race for the animals.…

1346 简单dp #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> using namespace std; #define N 200010 ]; int main() { int i,a,b; scanf("%d%d",&a,&am…

挺好的dp 因为有一点限制 必须任意去除一个数 总和就会小于另一个总和 换句话来说就是去除最小的满足 那么就都满足 所以是限制最小值的背包 刚开始从小到大定住最小值来背 TLE了一组数据 后来发现如果从大到小的话 就不用多加一重for了 前面算的已经记录下来 直接用就OK了 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib…