tjx41624560的博客

Problem You have been invited to the popular TV show “Would you like to be a millionaire?”.Of course you would! The rules of the show are simple: Before the game starts, the host spins a wheel of f

原题链接思路： 动态规划，回文符。 dp[j][k]是从j到k变为回文符所需的最小花费。AC代码：#include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib> #include <cstring> #include <cmath> using namespace std;int N,M; char s[

原文链接原题： Description Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labele

原题链接思路： 动态规划。 思路：i秒转移j次：dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]) + 第i秒是否接到苹果 dp[i-1][j]： i秒没有转移 dp[i-1][j-1]：i秒转移了 AC代码：#include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib

原题链接题目： n 组测试数据，有p个人在监狱，要放出q个人，这些人的编号为a[1]~a[q]，每放出一个人，他周围的人（两边连续的直到碰到空的监狱或者尽头）都要贿赂1枚金币，问最少花费多少金币。

Problem ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the park where n trees grow. They decided to be naughty and color the trees in the park. The trees are numbered with i

原题链接思路： 首先弄清楚T，A，S，B的意思。 T是蚂蚁有多少种。 A是蚂蚁有多少个。 每组蚂蚁数在S~B之间。 output有多少组符合条件。 然后用动态规划。 dp[i][j]在第i种蚂蚁前，已经取了j种蚂蚁。 dp[i+1][j+k] = dp[i][j] + dp[i][j+1] + …… + dp[i][j+k] （已经从i种之前选了j~j+k只蚂蚁，要再选取k~0只第

原题链接思路：动态规划。 动态方程：dp[i][j] = dp[i][j] + max(dp[i-1][j-1], dp[i-1][j]) dp[i][j]表示到点（i，j）的最大数字和。AC代码：#include <iostream>

原题链接 Description People in Silverland use coins.They have coins of value A1,A2,A3…An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice wa

原题链接思路： 动态规划。大数处理。高位是dp/mod，低位是dp%mod。AC代码：#include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib> #include <cstring> #include <cmath> using namespace std;//思路：参考完全背包问题。 //因为数很大，