挺水的一道题.规律性非常强,在数组中找出最大的数max,用max/m计算出倍数t,然后再把数组中的书都减去t*m,之后就把数组从后遍历找出第一个大于零的即可了 #include<iostream> #include<stdio.h> using namespace std; int main(){ // freopen("in.txt","r",stdin); int a[105],n,m; while(~scanf("%d%d&q…
题目链接:https://vjudge.net/problem/CodeForces-450B Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn modulo 1000000007 (109 + 7). Input The first line contains two integers x and y (|x|,…
题目链接:http://codeforces.com/problemset/problem/450/A /* * 计算一个人要是拿足够离开需要排多少次队,选排的次数多的那个人,如果两个人排的次数相同,那么要取后者: */ #include <cstdio> using namespace std; int main() { int n, m; int a, b, j, maxn; while (scanf("%d%d", &n, &m) != EOF) {…
链接:https://www.nowcoder.com/acm/contest/93/B来源:牛客网 题目描述 给你一个n*n矩阵,按照顺序填入1到n*n的数,例如n=5,该矩阵如下 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 现在让你连接相邻两条边的中点,然后只保留他们围成封闭图形区域的数字,那么这个矩阵变为 3 7 8 9 11 12 13 14 15 17 18 19 23 现在你们涵哥让你求变化后的矩…
题目:Click here 题意:给定数列满足求f(n)mod(1e9+7). 分析:规律题,找规律,特别注意负数取mod. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; ; int x, y, n; ]; int main() { while( ~scanf…
Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6012   Accepted: 3341 Description N children standing in circle who are numbered 1 through N clockwise are waiting their candies. Their teacher distributes the candies by in the following w…
题目链接: http://codeforces.com/problemset/problem/353/D?mobile=true H. Queue time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 There are n schoolchildren, boys and girls, lined up in the school canteen in front of the bun stall. The b…
Candy Distribution Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6033   Accepted: 3351 Description N children standing in circle who are numbered 1 through N clockwise are waiting their candies. Their teacher distributes the candies by…
这道题是有规律的博弈题目,,, 所以我们只需要找出规律来就ok了 牛人用sg函数暴力找规律,菜鸟手工模拟以求规律...[牢骚] if(m>=2) { if(n<=m) {first第一口就可以吃掉所有的.所以first必赢,} else {first无法一口吃掉所有的,所以second成了主动的了,如果first第一口吃掉k1个,那么明智的second只要吃掉k2个就可以了(n-k1-k2是偶数,也包括 0的),使得 剩下的数字是分成两个数字数目相等的堆,以后的工作便是first做什么,那么s…
已知有n个单位的水,问有几种方式把这些水喝完,每天至少喝1个单位的水,而且每天喝的水的单位为整数.看上去挺复杂要跑循环,但其实上,列举几种情况之后就会发现是找规律的题了= =都是2的n-1次方,而且这题输出二进制数就行了......那就更简单了,直接输出1,然后后面跟n-1个0就行了╮(╯_╰)╭ 下面AC代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm>…