Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6643    Accepted Submission(s): 2772 Problem Description The Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw t…
Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10883    Accepted Submission(s): 4610 Problem Description The Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw…
青蛙的约会 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions:132162   Accepted: 29199 Description 两只青蛙在网上相识了,它们聊得很开心,于是觉得很有必要见一面.它们很高兴地发现它们住在同一条纬度线上,于是它们约定各自朝西跳,直到碰面为止.可是它们出发之前忘记了一件很重要的事情,既没有问清楚对方的特征,也没有约定见面的具体位置.不过青蛙们都是很乐观的,它们觉得只要一直朝着某个方向跳下去,总能…
两只青蛙在网上相识了,它们聊得很开心,于是觉得很有必要见一面.它们很高兴地发现它们住在同一条纬度线上,于是它们约定各自朝西跳,直到碰面为止.可是它们出发之前忘记了一件很重要的事情,既没有问清楚对方的特征,也没有约定见面的具体位置.不过青蛙们都是很乐观的,它们觉得只要一直朝着某个方向跳下去,总能碰到对方的.但是除非这两只青蛙在同一时间跳到同一点上,不然是永远都不可能碰面的.为了帮助这两只乐观的青蛙,你被要求写一个程序来判断这两只青蛙是否能够碰面,会在什么时候碰面. 我们把这两只青蛙分别叫做青蛙A和…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2669 Romantic Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4179    Accepted Submission(s): 1745 Problem Description The Sky is Sprite.The Birds…
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead. InputThe input contains multiple test cases.Each case two nonnegative integer a,b (0<…
Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5489    Accepted Submission(s): 2164 Problem Description One day I was shopping in the supermarket. There was a cashier counting coins…
X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8416    Accepted Submission(s): 3066 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…
C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:33752   Accepted: 9832 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statement; I.e., a loop wh…
求解形如ax+by == n (a,b已知)的方程的非负整数解个数时,需要用到扩展欧几里得定理,先求出最小的x的值,然后通过处理剩下的区间长度即可得到答案. 放出模板: ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; } ll lcm(ll a, ll b) { return a / gcd(a,b) * b; } ll extend_gcd(ll a,ll b,ll&x,ll&y) { if(!b) { x = ; y = ; retur…