老顽童灬的博客

转载网址： 点击打开链接 点击打开题目 Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3495    Accepted Submission(s): 1381 Proble

矩阵相乘：#include<cstdio> int main() { int m1,m2,n1,n2; int t; int a[105][105]; int b[105][105]; int c[105][105]; scanf("%d",&t); while(t--) { scanf("%d%d",&m1,&n1);

Fat brother and Maze are playing a kind of special (hentai) game by two integers A and B. First Fat brother write an integer A on a white paper and then Maze start to change this integer. Every time M

点击打开链接 You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero

点击打开链接 Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0

Problem Description HOHO，终于从Speakless手上赢走了所有的糖果，是Gardon吃糖果时有个特殊的癖好，就是不喜欢将一样的糖果放在一起吃，喜欢先吃一种，下一次吃另一种，这样；可是Gardon不知道是否存在一种吃糖果的顺序使得他能把所有糖果都吃完？请你写个程序帮忙计算一下。   Input 第一行有一个整数T，接下来T组数据，每组数据占2

分解质因数： #include #include #include using namespace std; int pr(int n) { for(int i=2;i { if(n%i==0) { n/=i; printf("%d ",i); while(n%i==0) { n/=i; printf("%d ",i); } } } if(n>1)

转载网址：点击打开链接 最近见得博弈太多于是学习了一下 一个不错的PPT 传送门 巴什博弈 问题：只有一堆n个物品，两个人轮流从这堆物品中取物，规定每次至少取一个，最多取m个，最后取光者得胜。 公式：n%（m+1）！=0  当一个人要面对（m+1）的倍数时一定会输，另一个人只需一直让物品保持为n的倍数 减法博弈 巴什博弈有一种变形，叫做减法博弈。用s表示

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the sub

点击打开题目 给你一个传送门：点击打开链接 The Frog's Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 5707    Accepted Submission(s): 2748 Prob

#include<cstdio> #include<cstring> int zdgys(int a,int b)//辗转相除法 { int t; while(b) { t=b; b=a%b; a=t; } return t; } int zxgbs(int a,int b)//两个数的最小公倍数== 两个数相