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HDU 3555 Bomb
http://acm.hdu.edu.cn/showproblem.php?pid=3555
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7763 Accepted Submission(s): 2717
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?
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.思路:http://blog.csdn.net/winkloud/article/details/7866520
#include<iostream>#include<cstring>#include<cstdio>using namespace std;__int64 dp[25][3];int bit[25];void init(){ memset(dp,0,sizeof(dp)); int i; dp[0][0]=1; for(i=1;i<=20;i++) { dp[i][0]=dp[i-1][0]*10-dp[i-1][1]; //dp[i][0] 表示i位数字中不含49的数字的个数 dp[i][1]=dp[i-1][0]; //dp[i][1] 表示i位数字中以9开头的数字的个数 dp[i][2]=dp[i-1][2]*10+dp[i-1][1]; ////dp[i][2] 表示i位数字中含有49的数字的个数 }}__int64 solve(__int64 n){ memset(bit,0,sizeof(bit)); int len=1,i; __int64 temp=n; while(temp) { int r=temp%10; bit[len++]=r; temp=temp/10; } int flag=0; __int64 ans=0; bit[len]=0; for(i=len-1;i>=1;i--) { ans+=dp[i-1][2]*bit[i]; if(flag) { ans+=dp[i-1][0]*bit[i]; } //if(!flag&&bit[i+1]==4&&bit[i]>=9) //{ // ans+=dp[i][1]; // } if(!flag&&bit[i]>4) { ans+=dp[i-1][1]; } if(bit[i+1]==4&&bit[i]==9) flag=1; } return ans;}int main(){ int T; __int64 r; scanf("%d",&T); init(); while(T--) { scanf("%I64d",&r); printf("%I64d\n",solve(r+1)); } return 0;}
HDU 3555 Bomb
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