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
3150500
Sample Output
0115
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”,数位模板!!
但是脑残,,,,dp没开long long;
Code:
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include#include<vector>#include<string>#include<queue>#include<deque>#include<stack>#include<map>#include<set>#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;}</set></map></stack></deque></queue></string></vector></cmath></cstring></cstdio></iostream>