Description
Input
Output
Sample Input
3 1 50 500
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.
<pre name="code" class="plain">#include<iostream> #include<cstdio> #include<cstring> using namespace std; __int64 dp[22][4],n; int b[22]; //dp[i][0]表示长度为i,包括49的个数 //dp[i][1]表示长度为i,开头为9的个数 //dp[i][2]表示长度为i,没有49 void fun() { int i; memset(dp,0,sizeof(dp)); dp[0][2]=1; for(i=1;i<20;i++) { dp[i][0]=(__int64)dp[i-1][0]*10+dp[i-1][1];//有49=上一位有9的 + 上一位有 49 *10 dp[i][1]=dp[i-1][2];//有 9的 =上一位无 49的 dp[i][2]=(__int64)dp[i-1][2]*10-dp[i-1][1];// 没有49的 =上一位无 49*10-上一位有9的 } } int main() { int t,i; fun(); while(scanf("%d",&t)!=EOF) { while(t--) { scanf("%I64d",&n); n++; int len=0; while(n) { b[++len]=n%10; n/=10; } b[len+1]=0; bool f=0; __int64 ans=0; for(i=len;i>0;i--) { ans+=(__int64)b[i]*dp[i-1][0]; if(f) ans+=(__int64)dp[i-1][2]*b[i]; if(!f&&b[i]>4) ans+=dp[i-1][1]; if(b[i]==9&&b[i+1]==4) f=1; } printf("%I64d\n",ans); } } return 0; }
原文地址:http://blog.csdn.net/u012346225/article/details/42273493