HDU 3555 Bomb(数位DP)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 20202    Accepted Submission(s): 7548

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.
 

 

Author

fatboy_cw@WHU

 

Source

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

 

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zhouzeyong

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef __int64 LL;LL dp[27][3];int c[27];//dp[i][j]:长度为i的数的第j种状态//dp[i][0]:长度为i但是不包含49的方案数//dp[i][1]:长度为i且不含49但是以9开头的数字的方案数//dp[i][2]:长度为i且包含49的方案数void init(){    memset(dp,0,sizeof(dp));    dp[0][0] = 1;    for(int i = 1; i <= 20; i++)    {        dp[i][0] = dp[i-1][0]*10-dp[i-1][1];//所有情况减去开头为4的情况        dp[i][1] = dp[i-1][0]*1;//这个随便        dp[i][2] = dp[i-1][2]*10+dp[i-1][1];//已经有的随便，有9的就加上4就行    }}int cal(LL n){    int k = 0;    memset(c,0,sizeof(c));    while(n)    {        c[++k] = n%10;        n/=10;    }    c[k+1] = 0;    return k;}void solve(int len, LL n){    int flag = 0;                   //标记是否出现过49    LL ans = 0;    for(int i = len; i >= 1; i--)    {        ans+=c[i]*dp[i-1][2];       //为什么乘以c[i]因为可以有0~i-1个也就是i个        if(flag)                    //如果之前出现过49那么直接乘以没有出现49的情况，因为初始时候已经乘以了有49的情况            ans+=c[i]*dp[i-1][0];        else if(c[i] > 4)           //这一位前面没有挨着49，但c[i]比4大，那么当这一位填4的时候，要加上dp[i-1][1]            ans+=dp[i-1][1];        if(c[i+1]==4 && c[i]==9)            flag = 1;    }    printf("%I64d\n",ans);}int main(){    int t;    LL n;    init();    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        int len = cal(n+1);        solve(len, n);    }    return 0;}