ACM训练日志5 数位DP

   今晚花了俩小时做了一个题。题目如下：
https://cn.vjudge.net/contest/251878#problem/G
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.

题目大意：数出含有数字49的数字的个数。
题目WA了好几次，最终把所有的int都换成了long long ,终于AC了。。

AC代码：

#include <bits/stdc++.h>
using namespace std;
long long dp[20][10];
long long  bits[20];
long long dfs(int x, int y, bool end_flag)
{
    if(!end_flag&&dp[x][y]!=-1)
        return dp[x][y];
    if(x==0)return dp[x][y]=1;
    int end=end_flag?bits[x-1]:9;
    long long ans=0;
    for(long long i=0;i<=end;i++)
    {
        if(!(y==4&&i==9))
            ans+=dfs(x-1,i,end_flag&&(i==end));
    }
    if (!end_flag)
        dp[x][y]=ans;
    return ans;
}long long  solve(long long x)
{
    memset(bits,0,sizeof bits);
    int pos=0;
    while(x)
    {
        bits[pos++]=x%10;
        x/=10;
    }
    return dfs(pos, bits[pos], 1);
}
int main()
{
    memset(dp,-1,sizeof dp);
    int T;
    long long n;
    scanf("%d",&T);
    while(T--)
    {
       scanf("%lld",&n);
       printf("%lld\n",(n+1)-solve(n));
    }
    return 0;
}