HDU 4734 （数位DP）题解

思路：

dp[pos][pre]代表长度为pos的不大于pre的个数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<set>
#include<vector>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 100+5;
int dp[12][200000],a[20],UP;
int F(int x){
    int ret = 0,pos = 0;
    while(x){
        ret += (x % 10) *(1 << pos);
        pos++;
        x /= 10;
    }
    return ret;
}
int dfs(int pos,int sum,bool limit){
    if(pos == -1) return sum >= 0;
    if(sum < 0) return 0;
    if(!limit && dp[pos][sum] != -1) return dp[pos][sum];
    int top = limit? a[pos] : 9;
    int ret = 0;
    for(int i = 0;i <= top;i++){
        ret += dfs(pos-1,sum-i*(1<<pos),limit && i == top);
    }
    if(!limit) dp[pos][sum] = ret;
    return ret;
}
int solve(int x){
    int pos = 0;
    while(x){
        a[pos++] = x % 10;
        x /= 10;
    }
    return dfs(pos-1,UP,true);
}
int main(){
    int k = 1;
    memset(dp,-1,sizeof dp);
    int A,B;
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&A,&B);
        UP = F(A);
        printf("Case #%d: %d
",k++,solve(B));
    }
    return 0;
}