C. Dima and Salad   Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to have a walk, Inna decided to cook something. Dima and Seryozha have n fruits in the fridge. Each fruit has…

C. Dima and Salad time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to…

C. Dima and Salad time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to…

题目链接: http://codeforces.com/problemset/problem/272/D D. Dima and Two Sequences time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, …

D. Dima and Lisa Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/584/problem/D Description Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them…

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…

                                                 D. Dima and Lisa Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at mos…

#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…

题意: 给出一个矩阵n(<=500)*m(<=500)每一行任选一个数 异或在一起 求一个 异或在一起不为0 的每行的取值列号 思路: 异或的性质  交换律 x1^x2^x3==x3^x2^x1 可以任意换位置  并且 x1^x2==x3^x4  等于 x1^x2^x3==x4 可以任意换位置 所以等于零时有  x1^x2^x3^x4==0  (x1^x2)^(x3^x4)==0  x1^x2==x3^x4 都可以任意结合 所以本题只要任意选择列  如果为0  就每一列 找与已选择的列不一样的…