[動態規劃][數位dp]Bomb
阿新 • • 發佈:2018-08-06
sin input them ++ counter pac sam help while 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.
Sample Input
3
1
50
500
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
The input terminates by end of file marker.
Output For each test case, output an integer indicating the final points of the power.
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. 思路:數位dp--記憶化搜索 AC代碼:
#include <iostream> #include<cstdio> typedef long long ll; using namespace std; ll dp[15][2]; ll digit[15]; ll dfs(ll len,bool last_4,bool limit){ if(len==0) return 1; if(!limit && dp[len][last_4]) return dp[len][last_4]; ll sum=0; for(ll i=0;i<=(limit?digit[len]:9);i++){ if(!(last_4&&i==9)) sum+=dfs(len-1,i==4,limit&&i==digit[len]); } if(!limit) dp[len][last_4]=sum; return sum; } ll solve(ll x){ digit[0]=0; while(x){ digit[++digit[0]]=x%10; x/=10; } return dfs(digit[0],false,true); } int main() { ll t;scanf("%lld",&t); while(t--){ ll n;scanf("%lld",&n); printf("%lld\n",n-(solve(n)-1)); } return 0; }
[動態規劃][數位dp]Bomb