【HDU】3555【Bomb】

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题目链接

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 26679 Accepted Submission(s): 10102

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.

Author

fatboy_cw@WHU

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

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zhouzeyong

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题解

此题题意就是给你T个n，找出n以内的包含49的数的个数。

很裸的一题数位dp。

直接dp包含49的数的个数有点麻烦，所以我先算出不包含49的数的个数，然后用n+1来减（因为计算不包含49的数中是包括0的，所以减了之后还要加1）。

在搜索的过程中记录三个数，l表示当前是第几位，mach表示上一位是否为4，upp表示之前的位数是否都是取最大值，如果之前的数都是取最大值，当前的位只能取0到n的当前位，而不是0到9.

然后开始搜索，出现连续的49就跳过。

dp[j][i]记录第j位时上一位是否为4（i表示）不包含49的数的个数，用记忆化，如果搜过就直接加上。

因为上一位是否为4决定了当前位能否为9，对后面数的个数有影响，所以要加一维i。

上代码

#include<bits/stdc++.h>

using namespace std;

int t;

long long n;

int l,a[109];

long long dp[109][3];

long long dfs(int l,bool mach,bool upp) {

	if(l<=0) return 1;

	if(upp==0 && dp[l][mach]!=-1) return dp[l][mach];

	int up;

	if(upp==1) up=a[l];

	else up=9;

	long long sum=0;

	for(int j=0; j<=up; j++) {

		if(mach==1 && j==9) continue;

		sum+=dfs(l-1,j==4,upp==1 && j==up);

	}

	if(upp==0) dp[l][mach]=sum;

	return sum;

}

long long ans(long long x) {

	while(x>0) {

		l++;

		a[l]=x%10;

		x/=10;

	}

	return dfs(l,0,1);

}

int main() {

	scanf("%d",&t);

	memset(dp,-1,sizeof(dp));

	while(t--) {

		scanf("%lld",&n);

		l=0;

		printf("%lld\n",n+1-ans(n));

	}

	return 0;

}