HDU4734——F(x)(数位DP)

dp[i][j]表示i位数权值不超过j的数的个数

注意点

dp[i][j]的值不用每次都初始化,因为它的值不受输入的影响,如果前面算过了就直接拿来用,没算过就拿来算并记录下来

<strong>#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<cmath>
#include<cstdlib>
#include<cctype>
#define inf 0x3f3f3f3f
#define maxn 10000
typedef long long LL;
using namespace std;
int m,n,pos,f[11],fa;
int dp[11][5000];
int num[11];
int dfs(int pos,int s,int e)
{
    if(s<0) return 0;
    if(pos==-1) return s>=0;
    if(!e&&dp[pos][s]!=-1) return dp[pos][s];
    int end=e?num[pos]:9;
    int sum=0;
    for(int i=0;i<=end;i++){
        sum+=dfs(pos-1,s-i*f[pos],e&&i==end);
    }
    return e?sum:dp[pos][s]=sum;
}
int main()
{
    for(int i=0;i<=10;i++){
        f[i]=1<<i;
    }
    int t,iCase=1;
    cin>>t;
    memset(dp,-1,sizeof dp);
    while(t--){
        scanf("%d%d",&m,&n);
        int x=1,fa=0;
        while(m){
            fa+=m%10*x;
            x<<=1;
            m/=10;
        }
        pos=0;
        while(n){
            num[pos++]=n%10;
            n/=10;
        }
        printf("Case #%d: %d\n",iCase++,dfs(pos-1,fa,1));
    }
	return 0;
}</strong>

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时间: 2024-10-10 10:54:23

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