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                                                                       E. Dima and Kicks                                                              time limit per test 2 seconds                                                              memory li…
题目链接:http://codeforces.com/problemset/problem/358/D 开始题意理解错,整个就跪了= = 题目大意:从1到n的位置取数,取数的得到值与周围的数有没有取过有关,所有数都要取,求最终得到的最大结果 解题思路:dp题,转移方程如下 dp[i][0]=max(dp[i-1][0]+b[i-1],dp[i-1][1]+c[i-1]) dp[i][1]=max(dp[i-1][0]+a[i-1],dp[i-1][1]+b[i-1]) a,b,c分别表示周围没有…
#include <iostream> #include <algorithm> #include <vector> using namespace std; int check(int a, int b){ ); ) ; ; } int main(){ int n; cin >> n; vector<int> x(n); ; i < n; ++ i) cin >> x[i]; int i; ; i < n ; ++ i)…
#include <iostream> #include <algorithm> #include <string> using namespace std; int main(){ int n; cin >> n; string words = "<3",tmp; ; i < n ; ++ i){ cin >> tmp; words +=tmp+"<3"; } string messag…
A - Dima and Continuous Line 水题:直接模拟: #include<cstdio> #define maxn 1005 using namespace std; int x[maxn],y[maxn]; int main() { int n,a,b,last; scanf("%d",&n); ; ;i<n;i++) { scanf("%d",&b); a=last; last=b; )continue; i…
题面: 传送门 题目描述: 题意很简单:在一个N*M的矩阵中(N行M列),问是否可以:每行选一个整数,使他们的异或和大于0.如果不可以,输出"NIE":如果可以,输出"TAK",并输出选择的整数.   题目分析: 这道题刚开始想直接暴力,但看到复杂度竟然是O(500^500),就怂了.到后面仔细观察,发现题目有个数据:每个数小于1024,也就是2^10次方.会不会跟数位有关?后面分析了一下,果然是,但没时间做了,不过结束后还是A出来了.   异或(求大佬无视):转化…
A. Sereja and Dima time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja and Dima play a game. The rules of the game are very simple. The players have n cards in a row. Each card contains…
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…
Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…