HDU 3555 Bomb（数位DP）

　　　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　Bomb

　　　　　　　　　　　　　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　　　　　Total Submission(s): 11804    Accepted Submission(s): 4212

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.

 

 

Author

fatboy_cw@WHU

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

 

【思路】

       数位DP。

　　 预处理：

              f[i][0] 表示i位数中不含49的数的数目

              f[i][1] 表示i位数中不含49且最高位为9的数的数目

              f[i][2] 表示i位数中含49的数的数目

       根据n的每一位统计ans，具体见代码。

 

【代码】

 

#include<cstdio>
using namespace std;

typedef long long LL;
LL f[25][3];
/*
     f[i][0] i位数 无49存在
    f[i][1] i位数 无49存在且末尾为9
    f[i][2] i位数 有49
*/
int b[25];

void init() {
    f[0][0]=1; f[0][1]=f[0][2]=0;
    for(int i=1;i<25;i++) {
        f[i][0]=10*f[i-1][0]-f[i-1][1];
        f[i][1]=f[i-1][0];
        f[i][2]=10*f[i-1][2]+f[i-1][1];
    }
}

int main() {
    int T; LL n;
    scanf("%d",&T);
    init();
    while(T--) {
        scanf("%I64d",&n);
        int len=0;
        while(n)
            b[++len]=n%10 , n/=10;
        b[len+1]=0;
        LL ans=0; bool flag=0;
        for(int i=len;i;i--) {
            ans += b[i]*f[i-1][2];                    // + ...49... 
            if(flag) ans += f[i-1][0]*b[i];            // + 49...(无49...)
            else if(b[i]>4) ans += f[i-1][1];        // + 4(9 无49)
            if(b[i+1]==4 && b[i]==9) flag=1;
        }
        if(flag) ans++;                                //算上自身
        printf("%I64d\n",ans);
    }
    return 0;
}