hdu 4722 Good Numbers 数位dp

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cyendra

最新推荐文章于 2022-03-15 22:56:16 发布

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分类专栏： DP 解题报告

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本文详细介绍了如何使用数位DP技术解决给定区间内的数字，其各位数之和能被10整除的问题。通过将问题分解为子问题并利用记忆化搜索进行优化，提高了解决方案的效率。

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Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 602 Accepted Submission(s): 220

Problem Description

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10¹⁸).

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

Sample Input

2
1 10
1 20

Sample Output

Case #1: 0
Case #2: 1
Hint
The answer maybe very large, we recommend you to use long long instead of int.

---------------------

求区间[l,r]之间有多少个数，各位数和能被10整除。

转化为get(x),区间[0,x]之间有多少个数满足条件。

则答案即为get(r)-get(l-1)，问题转化为数位dp。

将x的各位数字储存在bit[]中。

考虑区间[0~5132]

可以转化成[0~999] 、[1000~1999]、[2000~2999]、[3000~3999]、[4000~4999]、[5000~5099]、[5100~5109]、[5110~5119]、[5120~5129]、[5130~5132]

令f[i][j]表示前i位和模10得j的方案数。若i为最后一位，若j=0则f[i][j]=1，否则f[i][j]=0。

用记忆化搜索实现数位dp

dp(int p,int m,bool flag)，flag=true 表示当前位p无论取多少都小于数x 。

则dp(i,m) += dp(i+1,(m+k)%10) 0=<k<= (9 or bit[i])

-------------------------

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;
typedef long long LL;
const int maxn=111;
LL f[maxn][11];
int bit[maxn];
int n;
LL dp(int p,int m,bool flag){
    if (p==0) return (m==0);
    if (flag&&f[p][m]!=-1) return f[p][m];
    int k=flag?9:bit[p];
    LL res=0;
    for (int i=0;i<=k;i++){
        res+=dp(p-1,(m+i)%10,flag||k!=i);
    }
    if (flag) f[p][m]=res;
    return res;
}

LL get(LL x){
    memset(f,-1,sizeof(f));
    if (x<0) return 0;
    n=0;
    while (x!=0){
        bit[++n]=x%10;
        x/=10;
    }
    return dp(n,0,0);
}

int main()
{
    int T;
    scanf("%d",&T);
    for (int cas=1;cas<=T;cas++){
        LL x,y;
        scanf("%I64d%I64d",&x,&y);
        printf("Case #%d: %I64d\n",cas,get(y)-get(x-1));
    }
    return 0;
}