HDOJ 题目3555 Bomb（数位dp）

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本文探讨了一个计数问题，即在从1到N的整数中寻找包含子序列“49”的数字个数。通过动态规划的方法，提供了一种有效解决此问题的算法实现。

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Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 8666 Accepted Submission(s): 3030

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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题目大意问一个数小于等于他出现过49的数的个数

dp[i][0]代表的是出现的个数

dp[i][1]代表没有出现但是最高位的9的个数

dp[i][2]代表的是没有出现的个数

ac代码

#include<stdio.h>
#include<string.h>
__int64 dp[25][3];
int bit[25];
void init()
{
	int i;
	memset(dp,0,sizeof(dp));
	dp[0][2]=1;
	for(i=1;i<25;i++)
	{
		dp[i][0]=dp[i-1][0]*(__int64)10+dp[i-1][1];//当已经出现49，这一位就是10个数随便填，或者上一位是9这一位填4就好
		dp[i][1]=dp[i-1][2];//就在没出现的基础上前边填9就好
		dp[i][2]=dp[i-1][2]*(__int64)10-dp[i-1][1];//以前没出现在最前边填任意10个数再减去上一位为9这一位是4的情况
	}
}
__int64 slove(__int64 n)
{
	int i,len=0,flag=0;
	__int64 ans=0;
	memset(bit,0,sizeof(bit));
	n++;
	while(n)
	{
		bit[++len]=n%10;
		n/=10;
	}
	bit[len+1]=0;
	for(i=len;i;i--)
	{
		ans+=dp[i-1][0]*(__int64)bit[i];
		if(flag)
		{
			ans+=dp[i-1][2]*(__int64)bit[i];
		}
		if(!flag&&bit[i]>4)
		{
			ans+=dp[i-1][1];
		}
		if(bit[i]==9&&bit[i+1]==4)
		{
			flag=1;
		}
	}
	return ans;
}
int main()
{
	int t;
	scanf("%d",&t);
	init();
	while(t--)
	{
		__int64 n;
		int len=0;
		scanf("%I64d",&n);
		printf("%I64d\n",slove(n));
	}
}