Bomb

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/65536 K (Java/Others)

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

500

Sample Output

题意：

问[1,n]中有多少含49的数;

思路：

位数dp;

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e6+5e5;
LL dp[25][3],n;
int b[25];
int fun()
{
    mst(dp,0);
    dp[0][2]=1LL;
    for(int i=1;i<23;i++)
    {
        dp[i][0]=dp[i-1][0]*10+dp[i-1][1];
        dp[i][1]=dp[i-1][2];
        dp[i][2]=dp[i-1][2]*10-dp[i-1][1];
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    fun();
    while(t--)
    {
        scanf("%lld",&n);
        LL temp=n,ans=0;
        int cnt=1;
        while(temp)
        {
            b[cnt++]=temp%10;
            temp/=10;
        }
        b[cnt]=0;
        int flag=0;
        for(int i=cnt;i>0;i--)
        {
            ans+=dp[i-1][0]*(LL)b[i];
            if(b[i]>4&&!flag)
            {
                ans+=dp[i-1][1];
            }
            if(flag)
            {
                ans+=dp[i-1][2]*b[i];
            }
            if(b[i+1]==4&&b[i]==9)
            {
                flag=1;
            }
        }
        if(flag)ans++;
        printf("%lld
",ans);
    }
    return 0;
}