hdu 4722 Good Numbers(数位dp)

最新推荐文章于 2022-03-15 22:56:16 发布

转载最新推荐文章于 2022-03-15 22:56:16 发布 · 61 阅读

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#java #数据结构与算法

本文介绍了一种使用数位DP方法解决特定好数字计数问题的算法。该问题要求找出两个给定数字范围内的好数字数量，其中好数字定义为该数字的所有位数之和能被10整除的数字。文章提供了完整的C++代码实现，并对关键函数进行了注释说明。

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Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3453 Accepted Submission(s): 1090

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 <= 10 ¹⁸).

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.

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

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zhuyuanchen520

题意：求A到B之间各个位数的和sum%10==0 的数的个数。

题解：数位dp,dp[i][j]表示当前i位mod10=j的个数。

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#define ll long long

using namespace std;

int n;

ll dp[20][20];
int num[20];
ll a,b;

ll dfs(int i,int mod,bool e) {
    if(i<=0)return mod?0:1;
    if(!e&&dp[i][mod]!=-1)return dp[i][mod];
    ll res=0;
    int u=e?num[i]:9;
    for(int d=0; d<=u; d++) {
        int Mod=(mod+d)%10;
        res+=dfs(i-1,Mod,e&&d==u);
    }
    return e?res:dp[i][mod]=res;
}

ll solve(ll x) {
    int len=1;
    ll k=x;
    while(k) {
        num[len++]=k%10;
        k/=10;
    }
    num[len]=0;
    return dfs(len-1,0,1);
}

int main() {
    //freopen("test.in","r",stdin);
    int t;
    memset(dp,-1,sizeof dp);
    cin>>t;
    int ca=1;
    while(t--) {
        scanf("%I64d%I64d",&a,&b);
        printf("Case #%d: %I64d\n",ca++,solve(b)-solve(a-1));
    }
    return 0;
}