HDU 3555 数位dp

最新推荐文章于 2019-01-16 13:07:32 发布

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最新推荐文章于 2019-01-16 13:07:32 发布

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本文介绍了一道编程题目，要求计算从1到N内包含子串49的数字数量。通过使用数位DP的方法，提供了一个有效的解决方案，并附带了完整的代码实现。

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

Sample Output

0
1
15

Hint
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.

题意：给你一个n，求1-n中包含49这个子串的数有多少。

题解：数位dp。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
ll digit[20],dp[20][20][20][20];
ll dfs(ll have,ll pos,ll flag,ll lim){
	if(pos<=0){
		return have;
	}
	if(!lim&&dp[have][pos][flag][lim]!=-1){
		return dp[have][pos][flag][lim];
	}
	ll num=lim?digit[pos]:9;
	ll i,ret=0;
	for(i=0;i<=num;i++){
		if(flag==4&&i==9)ret+=dfs(1,pos-1,i,lim&&(i==num));//如果当前等于4且这一位为9，那就代表有了49这个子串
		else ret+=dfs(have,pos-1,i,lim&&(i==num));
	}
	if(!lim)dp[have][pos][flag][lim]=ret;
	return ret;
}
ll solve(ll n){
	ll len=0;
	while(n){
		digit[++len]=n%10;
		n/=10;
	}
	ll ret=0,i;
	digit[len+1]=0;
	return dfs(0,len,0,1);
}
int main(){
	ll t;
	memset(dp,-1,sizeof(dp));
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		cout<<solve(n)<<endl;
	}
	return 0;
}