题目传送门 /* 递推DP: dp[i] 表示放i的方案数,最后累加前n-2的数字的方案数 */ #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; ]; int main(void) //URAL 1260 Nudnik Photographer { //fr…

题目链接 题意 : 给你1到n这n个数,排成一排,然后1放在左边最开始,剩下的数进行排列,要求排列出来的数列必须满足任何两个相邻的数之间的差不能超过2,问你有多少种排列 思路 : 对于dp[n], n个人时求F[n].第2个位置放2时有F[n-1]种:第2个位置放3,第3个位置放2,第4个位置只能放4,有F[n-3]种:第2个位置放3,第3个位置放5,13578642,有1种:第2个位置放3,第3个位置不能放4. 所以: 1.12……(dp[n-1]) 2.1324……(dp[n-3]) 3.1…

Problem Description If two people were born one after another with one second difference and one of them is a child, then the other one is a child too. We get by induction that all the people are children. Everyone knows that the mathematical departm…

题目:click here :这个题可以先dfs深搜下,规律dp dfs: #include <bits/stdc++.h> using namespace std; #define S second typedef long long ll; const int INF = 0x3f3f3f3f; ; int s; ; ]; void dfs( int last, int num ) { if( num == s ) { cnt++; return ; } ; i<s+; i++ )…

A nudnik photographer 大意: 对1到N这些数进行排列,1必需要在最左边.相邻的两个数之间的差值不能超过2,问有多少种排列的方法. 思路: 对座位进行DP,当第一个是1,第二个是2的时候,组合为dp[i-1].当第一个是1,第二个是3的时候,第三个也确定了是2.组合为dp[i-3]:还有最后一种情况是1357--8642. 所以DP方程为dp[i] = dp[i-1]+dp[i-3]+1. #include <stdio.h> int n; int dp[100]; int…

Time limit: 1.0 second Memory limit: 64 MB If two people were born one after another with one second difference and one of them is a child, then the other one is a child too. We get by induction that all the people are children. Everyone knows that t…

列表: URAL 1225 Flags URAL 1009 K-based Numbers URAL 1119 Metro URAL 1146 Maximum Sum URAL 1203 Scientific Conference URAL 1353 Milliard Vasya's Function URAL 1260 Nudnik Photographer URAL 1012 K-based Numbers. Version 2 URAL 1073 Square Country URAL 1…

这几天扫了一下URAL上面简单的DP 第一题 简单递推 1225. Flags #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define LL long long LL dp[][]; int main() { int i,n; scanf("%d",&n); dp[][] = ;d…

HDU 1260 Tickets 题意:有N个人要买票,你可以一个一个人卖票,时间分别为Xs,也可以相邻两个人一起卖票,时间为Ys,从早上八点开始卖票,问你何时最早将N个人的票卖完. 思路:解决情况是当前最优,要么就单卖,状态最优就是前一个人的,要么和前一个人一起拼凑,状态最优是前两个人的,取时间最短的. 时间显示问题,注意下上下午的事情就好了. dp[i] = min(dp[i-1]+signal[i],dp[i-2]+unit[i]) /** Sample Input 2 2 20 25 4…

题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=1260 分析: f[i][j]表示i~j刷成s[i]~s[j]这个样子需要的最小次数 则若s[i]==s[j]:f[i][j]=min(f[i+1][j],f[i][j-1],f[i+1][j-1]+1) 若s[i]!=s[j]:f[i][j]=min(f[i][k]+f[k+1][j])…