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杭电 3555 Bomb

2024-07-12 00:39:11   220人阅读

Bomb


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6609    Accepted Submission(s): 2303




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
3
1
50
500
 


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.



第一个数位动规！！！！

看代码理解的，差不多懂那么点点点了。


AC代码如下：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;


int main()
{
    __int64 dp[20][3];//0表示该位数中不含49的情况数，1表示不含49但头为9的情况数，2表示含49的情况数
    int i,j;
    memset(dp,0,sizeof dp);
    dp[0][0]=1;
    for(i=1;i<=20;i++)
    {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
        dp[i][1]=dp[i-1][0];
        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
        //cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl;
    }
    int t;
    cin>>t;
    __int64 num[30];
    __int64 n;
    while(t--)
    {
        memset(num,0,sizeof num);
        cin>>n;
        n++;//这一步很重要
        int tt=1;
        while(n)
        {
            num[tt++]=n%10;
            n/=10;
        }
        __int64 ans=0;
        int flag=0;int last=0;
        for(i=tt-1;i>=1;i--)
        {
            ans+=num[i]*dp[i-1][2];
            if(flag==1)
                ans+=dp[i-1][0]*num[i];
            if(flag==0&&num[i]>4)
                ans+=dp[i-1][1];
            if(num[i+1]==4&&num[i]==9)
                flag=1;
        }
        cout<<ans<<endl;
    }
    return 0;
}



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