杭电3555

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

3150500

 

Sample Output

0115
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.
 

做省赛题遇到数位dp就想学习一下

研究了很多天

感谢网上大神思路

题目大意 求出1---2^63-1中数字里含有49的数字的个数

#include<iostream>#include<cstdio>#include<queue>#include<cstring>#include<cmath>#include<cstdlib>#define mod 1000000007using namespace std;long long dp[21][3]={1};long long pow(long long i){//求出10的i次方    long long sum=1;    while(i--)sum*=10;    return sum;}int main(){    for(int i=1;i<=20;i++){//        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//到第i位不含49的数的个数         dp[i][1]=dp[i-1][0];//到第i位不含49但是位数是9的数的个数        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//到第i位含49的数的个数    }    int t;    cin>>t;    while(t--){        long long n,a[30],len=0;        cin>>n;        long long nn=n;        while(n){            a[++len]=n%10;            n/=10;        }        long long k=len;        long long sum=0,before=0;        while(len){            sum+=dp[len-1][2]*a[len];            if(a[len]>4)                sum+=dp[len-1][1];//这里好好理解            if(a[len]==9&&before==4){                sum+=(nn%pow(len-1));                sum++;                break;            }            before=a[len];            len--;        }        cout<<sum<<endl;    }    return 0;}