4421: [Cerc2015] Digit Division Time Limit: 1 Sec  Memory Limit: 512 MBSubmit: 348  Solved: 202[Submit][Status][Discuss] Description 给出一个数字串,现将其分成一个或多个子串,要求分出来的每个子串能Mod M等于0. 将方案数(mod 10^9+7)   Input 给出N,M,其中1<=N<=300 000,1<=M<=1000 000. 接下来一行…

4421: [Cerc2015] Digit Division 题目连接: http://www.lydsy.com/JudgeOnline/problem.php?id=4421 Description 给出一个数字串,现将其分成一个或多个子串,要求分出来的每个子串能Mod M等于0. 将方案数(mod 10^9+7) Input 给出N,M,其中1<=N<=300 000,1<=M<=1000 000. 接下来一行,一个数字串,长度为N. Output 如题 Sample In…

传送门 解题思路 差点写树套树...可以发现如果几个数都能被\(m\)整除,那么这几个数拼起来也能被\(m\)整除.同理,如果一个数不能被\(m\)整除,那么它无论如何拆,都无法拆成若干个可以被\(m\)整除的数.这样的话只需要看那些被\(m\)整除的前缀个数,然后选与不选直接\(2^cnt\)即可. 代码 #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #inc…

[BZOJ4421][Cerc2015] Digit Division Description 给出一个数字串,现将其分成一个或多个子串,要求分出来的每个子串能Mod M等于0. 将方案数(mod 10^9+7) Input 给出N,M,其中1<=N<=300 000,1<=M<=1000 000. 接下来一行,一个数字串,长度为N. Output 如题 Sample Input 4 2 1246 Sample Output 4 题解:如果一个前缀a%m==0,另一个长一点的前缀b…

如果两个相邻的串可行,那么它们合并后一定可行,所以求出所有可行的串的个数$t$,则$ans=2^{t-1}$. 注意特判整个串不可行的情况,这个时候答案为0. #include<cstdio> int n,m,i,t,ans;char a[300010]; int main(){ for(scanf("%d%d%s",&n,&m,a);i<n;i++){ t=(t*10+a[i]-'0')%m; if(!t)if(!ans)ans=1;else ans…

题目描述 We are given a sequence of n decimal digits. The sequence needs to be partitioned into one or more contiguous subsequences such that each subsequence, when interpreted as a decimal number, is divisible by a given integer m. Find the number of di…

Digit Division 题目链接: http://acm.hust.edu.cn/vjudge/contest/127407#problem/D Description We are given a sequence of n decimal digits. The sequence needs to be partitioned into one or more contiguous subsequences such that each subsequence, when interp…

Digit Division Time limit: 1 s Memory limit: 512 MiB We are given a sequence of n decimal digits. The sequence needs to be partitioned into one or more contiguous subsequences such that each subsequence, when interpreted as a decimal number, is divis…

题目链接:Central Europe Regional Contest 2015 Zagreb, November 13-15, 2015 D.Digit Division(排列组合+思维) 题解:如果这个数从划分的过程中第一.二道竖线前的能够整除m,那么第一道与第二道竖线之间的数也能够整除m. #include <bits/stdc++.h> using namespace std; typedef long long ll; char a[300005]; ll qpow(ll x,ll…

[Cerc2015]Kernel Knights Time Limit: 2 Sec Memory Limit: 512 MBSubmit: 5 Solved: 4[Submit][Status][Discuss]Description “Jousting”是一种让骑士在高速骑行中用木制长矛相互攻击对方的中世纪竞技游戏.现在,一共有2n个骑士进入一场“Jousting”锦标赛.骑士们被平均分配到2个house.竞赛开始时,所有骑士都会对另一个house的骑士之一发起挑战.一组解被定义为一个集合S…