HDU3555 Bomb

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 19868    Accepted Submission(s): 7384

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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题目的意思是求小于等于n的包含49的数有几个

思路：数位dp，三维数组保存到len为止，结尾是i的，是否已经有49的数有几个，dfs维护

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 10000005#define Mod 10001using namespace std;#define LL long longLL dp[200][20][2];int a[10];LL dfs(int len,int pre,int sta,bool limit){    if(len<0)        return sta;    if(dp[len][pre][sta]!=-1&&!limit)        return dp[len][pre][sta];    int up=limit?a[len]:9;    LL ans=0;    for(int i=0; i<=up; i++)    {        ans+=dfs(len-1,i,(pre==4&&i==9)||sta,limit&&i==up);    }    return limit?ans:dp[len][pre][sta]=ans;}LL solve(LL x){    memset(dp,-1,sizeof dp);    int cnt=0;    while(x>0)    {        a[cnt++]=x%10;        x/=10;    }    return dfs(cnt-1,-1,0,1);}int main(){    int T;    LL n;    for(scanf("%d",&T); T--;)    {        scanf("%lld",&n);        printf("%lld\n",solve(n));    }    return 0;}