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?
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.
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
题意：给你一个数n，数出[1,n]中含有49的个数。
思路：可以先把[0,r)中不含49的个数求出来，然后用n+1-solve(n+1)就行了。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
#define inf 99999999
#define pi acos(-1.0)
#define maxn 1000050
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef long double ldb;
ll dp[65][12];
void init()
{
int i,j,k;
memset(dp,0,sizeof(dp));
for(j=0;j<=9;j++)dp[1][j]=1;
for(i=2;i<=64;i++){
for(j=0;j<=9;j++){
for(k=0;k<=9;k++){
if(j==4 && k==9 )continue;
dp[i][j]+=dp[i-1][k];
}
}
}
}
ll solve(ll x)
{
int wei[70],i,j,len=0;
ll t=x;
while(t){
wei[++len]=t%10;
t/=10;
}
wei[len+1]=0;
ll sum=0;
for(i=len;i>=1;i--){
for(j=0;j<wei[i];j++){
sum+=dp[i][j];
}
if(wei[i+1]==4 && wei[i]==9)break;
}
return sum;
}
int main()
{
int m,i,j,T;
ll n;
init();
scanf("%d",&T);
while(T--)
{
scanf("%lld",&n);
printf("%lld
",n+1-solve(n+1));
}
return 0;
}