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Bomb
Bomb
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?
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.
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
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.
题意:求1到n一共有多少个49
一个简单的数位DP题。
<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; }
Bomb
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