题意:已知a,b,求的最后一位. 分析: 1.若b-a>=5,则尾数一定为0,因为连续5个数的尾数要么同时包括一个5和一个偶数,要么包括一个0. 2.若b-a<5,直接暴力求即可. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstre…

B. The Eternal Immortality time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise ag…

B. The Eternal Immortality 题目链接http://codeforces.com/contest/869/problem/B 解题心得:题意就是给出a,b,问(a!)/(b!)的个位数,要注意0,5两个数,只要a,b相差超过5个位数就只能是0,其实没有看到相差5看到相差10也可以的,然后又暴力跑一个末位数就可以了. /*这里跑的是相差10位*/ #include<bits/stdc++.h> using namespace std; typedef long long…

[链接] 链接 [题意] 求b!/a!的最后一位数字 [题解] b-a>=20的话 a+1..b之间肯定有因子2和因子5 答案一定是0 否则暴力就好 [错的次数] 在这里输入错的次数 [反思] 暴力很大,但是差值很小就确定了. 可以作为一个trick [代码] #include <bits/stdc++.h> #define ll long long using namespace std; ll a,b; int main() { //freopen("F:\\rush.tx…

Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lif…

Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lif…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ash…

Codeforces Round #439 (Div. 2) codeforces 869 A. The Artful Expedient 看不透( #include<cstdio> int main(){ puts("Karen"); ; } 15ms codeforces 869B. The Eternal Immortality(数学,水) 题意:输出两个数的阶乘的商的 个位数 题解:两数之差大于5,个位数就是0.小于5直接个位相乘即可. #include<cs…

A. The Artful Expedient 题目链接:http://codeforces.com/contest/869/problem/A 题目意思:给你两个数列,各包含n个数,现在让你从上下两个数列中各取一个数a[i],b[j],如果a[i]^b[j]在这2×n个数里面出现过,那么就获得一分,问将任意的a[i],b[j]之间的亦或之后,如果分数是奇数则Koyomi胜利,否则Karen胜利.问最后到底谁胜了. 题目思路:非常无聊的题目,暴力都可以过,就是暴力枚举a[i],b[j],把所有答…

A. The Artful Expedient time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her broth…