hdu 3555 Bomb（数位dp）

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 9343 Accepted Submission(s): 3303



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

HintFrom 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


含49的个数。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#define N 1010
#define ll long long

using namespace std;

ll dp[30][3];//0:合法  1:9开头  2：不合法 
             ///dp[i][j]  长度为i的j状态

void init() {
    memset(dp,0,sizeof dp);
    dp[0][2]=1;
    for(int i=1; i<=24; i++) {
        dp[i][0]=dp[i-1][0]*10+dp[i-1][1];
        dp[i][1]=dp[i-1][2];
        dp[i][2]=dp[i-1][2]*10-dp[i-1][1];
    }
}

ll solve(ll n) {
    int a[40];
    int len=1;
    while(n) {
        a[len++]=n%10;
        n/=10;
    }
    a[len]=0;
    ll ans=0;
    int flag=0;
    for(int i=len-1; i>=1; i--) {
        ans+=dp[i-1][0]*a[i];
        if(flag)
            ans+=dp[i-1][2]*a[i];
        if(!flag&&a[i]>4)
            ans+=dp[i-1][1];
        if(a[i]==9&&a[i+1]==4)  ///xx49***的情况  *为待填数位
            flag=1;
    }
    return ans;
}

int main() {
    init();
    int t;
    cin>>t;
    while(t--) {
        ll n;
        cin>>n;
        printf("%I64d\n",solve(n+1));
    }
    return 0;
}