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HDU 3555 Bomb [數位]

編輯：關於C++

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

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.

Analyse:

同 “不要62”，數位模板！！

但是腦殘，，，，dp沒開long long；

Code：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x7fffffff
#define SUP 0x80000000
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

typedef __int64 LL;
const int N=100007;
LL dp[30][10][2];
int digit[30];

LL dfs(int pos,int pre,int en,int limit)
{
    if(pos==-1) return en;
    if(!limit&&dp[pos][pre][en]!=-1) return dp[pos][pre][en];
    int last=limit?digit[pos]:9;
    LL ret=0;
    for(int i=0;i<=last;i++)
    {
        ret+=dfs(pos-1,i,en||(pre==4&&i==9),limit&&i==last);
    }
    if(!limit) dp[pos][pre][en]=ret;
    return ret;
}

LL solve(LL x)
{
    int cnt=0;
    while(x)
    {
        digit[cnt++]=x%10;
        x/=10;
    }
    return dfs(cnt-1,0,0,1);
}

int main()
{
    int T;scanf("%d",&T);
    LL n;
    mem(dp,-1);
    while(T--)
    {
        scanf("%I64d",&n);
        printf("%I64d\n",solve(n));
    }
    return 0;
}