Bomb
HDU - 3555
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.
Output For each test case, output an integer indicating the final points of the power. Sample Input
3 1 50 500Sample 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.
初始化不要忘
code：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
ll dp[30][10][2];//dp[pos][pre][yes]
int digit[30];
ll dfs(int pos,int pre,int yes,int limit){
if(pos == -1) return yes;
ll &dpnow = dp[pos][pre][yes];
if(!limit && dpnow != -1) return dpnow;
int max_digit = limit ? digit[pos] : 9;
ll ans = 0;
for(int i = 0; i <= max_digit; i++){
if(yes)
ans += dfs(pos - 1, i, 1, limit && i == max_digit);
else
ans += dfs(pos - 1, i, pre == 4 && i == 9, limit && i == max_digit);
}
if(!limit)
dpnow = ans;
return ans;
}
ll solve(ll n){
int pos = 0;
while(n){
digit[pos++] = n % 10;
n /= 10;
}
return dfs(pos-1,0,0,1);
}
int main(){
int t;
scanf("%d",&t);
memset(dp,-1,sizeof(dp));
while(t--){
ll n;
scanf("%lld",&n);
printf("%lld\n",solve(n));
}
return 0;
}