HDU_4734_数位dp

http://acm.hdu.edu.cn/showproblem.php?pid=4734

模版题。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int dp[][],a,b,num[];

int dfs(int sum,int pos,int limit)
{
    if(pos == )    return sum >= ;
    if(sum < )     return ;
    if(!limit && dp[sum][pos] != -)    return dp[sum][pos];
    int ans = ,endd = limit?num[pos]:;
    for(int i = ;i <= endd;i++)
    {
        ans += dfs(sum-i*(<<(pos-)),pos-,limit && i == endd);
    }
    if(!limit)  dp[sum][pos] = ans;
    return ans;
}
int main()
{
    memset(dp,-,sizeof(dp));
    int T;
    scanf("%d",&T);
    for(int i = ;i <= T;i++)
    {
        scanf("%d%d",&a,&b);
        int sum = ,now = ;
        while(a)
        {
            sum += a%*now;
            now *= ;
            a /= ;
        }
        now = ;
        while(b)
        {
            num[now] = b%;
            now++;
            b /= ;
        }
        printf("Case #%d: %d\n",i,dfs(sum,now-,));
    }
    return ;
}