HDU3555----Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 4594 Accepted Submission(s): 1601 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
3 1 50 500
Sample Output
0 1 15
传送门
题意：求1~N中含有数字49的个数 1 <= N <= 2^63-1 #include <cstdio> #include <cstring> using namespace std; int bit[25]; __int64 dp[25][3]; //dp[i][0]表示长度为i,没有49 //dp[i][1]表示长度为i,没有49但前一位为4 //dp[i][2]表示长度为i,包括49的个数 /*limit表示是否有上限,比如n=1234,现在转移到12,如果下一位选3,那么再下一位就有上限, 上限为4,如果不选3,那么下一位就没限制,最高位9,转移能保证转移到数比n小*/ __int64 Dfs (int pos,int s,bool limit) //s为之前数字的状态 { if (pos==-1) return s==2; if (limit==false && ~dp[pos][s]) return dp[pos][s]; int i ,end=limit?bit[pos]:9; __int64 ans=0; for (i=0;i<=end;i++) { int nows=s; if(s==0 && i==4) nows=1; if(s==1 && i!=9) //前一位为4 nows=0; if(s==1 && i==4) nows=1; if(s==1 && i==9) //49 nows=2; ans+=Dfs(pos-1 , nows , limit && i==end); } //limit==true则说明有限制,即所有可能并没有被全部记录,故此时记入dp数组 //limit==false则说明之后的分支状态已经搜索完全 return limit?ans:dp[pos][s]=ans; } int main () { __int64 n; int T; memset(dp,-1,sizeof(dp)); scanf("%d",&T); while (T--) { scanf("%I64d",&n); int len=0; while (n) { bit[len++]=n%10; n/=10; } printf("%I64d\n",Dfs(len-1,0,1)); } return 0; }
