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最新推荐文章于 2021-07-08 09:39:09 发布

顺顺不吃竹笋

最新推荐文章于 2021-07-08 09:39:09 发布

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本文链接：https://blog.youkuaiyun.com/DW_css/article/details/79216942

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第五次入围！——我校晋级第42届ACM国际大学生程序设计竞赛全球总决赛

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 21085 Accepted Submission(s): 7890

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

第一次碰到这种题，第一反应是暴力，，，，后来在百度找到了一篇比较清楚的文章

http://www.bubuko.com/infodetail-302143.html

假设输入的t=235，那么分四次统计：

1~199 确定2这一位

200~229 确定3这一位

230~234 确定4这一位

235 还剩一个本身的数235，最后判断自身是不是

#include<iostream>
#include<cstdio>
//#define ll long long
typedef long long ll;
using namespace std;

ll dp[25][3];           //这里一开始忘记改成ll，一直wa，心累 
/*
*有三种状态:
*第一是 dp[i][0]，在长度为i的数字里不存在为49的数字； 
*第二是 dp[i][1]，在长度为i的数字不含49但是最高位是9（只要高位在加4就是49XXXXX） 
*第三是 dp[i][2]，在长度为i的数字里存在数字是49的 
*/
int bit[21];
ll t;


void init()
{
	dp[0][0] = 1;
	dp[0][1] = 0;
	dp[0][2] = 0;
	for(int i = 1;i <= 21;++i)
	{
		dp[i][0] = dp[i-1][0]*10 - dp[i-1][1];        //除去49XXX的情况 
		dp[i][1] = dp[i-1][0];                        //前一段长度的不含49的数字个数前面直接加上9 
		dp[i][2] = dp[i-1][1] + dp[i-1][2]*10;        // 前一段长度最高位为9的直接加上4以及前一段有49的 
	}
}

ll cal(ll n)
{
	int temp = 0;
	while(n)
	{
		bit[++temp] = n%10;
		n /= 10;
	}
	ll ans = 0;
	bit[temp+1] = 0;
	bool has = false;
	for(int i = temp;i >= 1;--i)
	{
		   //bit[i]指的是有几个数字，假如bit[i]=5，那么该位上随便0,1,2,3,4都可以 
		ans += dp[i-1][2]*bit[i];          
		if(has)
		{
			//这里我还没搞懂┭┮﹏┭┮ 
			ans += dp[i-1][0]*bit[i]; 
		}
		else
		{
			if(bit[i]>4)
			{
				ans += dp[i-1][1];              //低位是9的，前面加上4 
			}
		}
		if(bit[i+1]==4 && bit[i]==9)
		{
			has = true;
		} 
	}
	if(has)
	{
		ans++;
	}
	return ans;
}

int main()
{
	init(); 
	int n;
	scanf("%d",&n);
	while(n--)
	{	
		scanf("%lld",&t); 
		printf("%lld\n",cal(t));
	}
	return 0;
 }