Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn‘t think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
The first line contains two integers: n (1?≤?n?<?1018) and m (1?≤?m?≤?100).
In a single line print a single integer — the number of numbers close to number n modulo m.
104 2
3
223 4
1
7067678 8
47
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
好久没做cf了，突然打开一道题，看别人写法瞬间眼睛一亮，第二次认识hash的写法。
/* ***********************************************
Author :rabbit
Created Time :2014/3/16 13:59:04
File Name :3.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
ll dp[100010][150],num[15],base[15],val[15];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
ll n,m;
while(cin>>n>>m){
memset(num,0,sizeof(num));
memset(base,0,sizeof(base));
memset(val,0,sizeof(val));
while(n){
num[n%10]++;
n/=10;
}
val[0]=1;
for(int i=0;i<10;i++){
base[i]=num[i]+1;
val[i+1]=base[i]*val[i];
}
dp[0][0]=1;
for(ll i=0;i<val[10];i++)
for(ll j=0;j<10;j++){
if(j==0&&i<val[1])continue;
if(i/val[j]%base[j]>=num[j])continue;
for(ll k=0;k<m;k++)
dp[i+val[j]][(k*10+j)%m]+=dp[i][k];
}
cout<<dp[val[10]-1][0]<<endl;
}
return 0;
}
CF 401D 状态压缩或者hashdp,布布扣,bubuko.com
原文：http://blog.csdn.net/xianxingwuguan1/article/details/21324915