HDU 3555 Bomb [数位DP]【动态规划】
2016-08-22 20:34
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题目链接 :HDU最近比较炸 不贴链接了、、
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Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15374 Accepted Submission(s): 5565
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
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.
———————————-.
题目大意 : 就是问你从1~m 里 有49的数的个数
解题思路 :
最最基础的数位DP
附本题代码
——————————————————.
——————.
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15374 Accepted Submission(s): 5565
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
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.
———————————-.
题目大意 : 就是问你从1~m 里 有49的数的个数
解题思路 :
最最基础的数位DP
附本题代码
——————————————————.
#include <bits/stdc++.h> using namespace std; #define LL long long int LL dp[30][12][2]; // 位数 数字0~9 状态 int num[30],len ; // 数字 长度 /**********记忆化搜索****************/ //分别代表现在的位数 上一位的数字 限制 状态 /* 主要就是限制 比如说m = 123456789; 那么位数到地5位的时候 就不能选6 7 ..等等了 但是如果在第5位他选择了 4 那么下一位就能够任意选择了 、 */ /* 状态就是表示 到现在 有没有他能够满足的状态 有就+1 没有就+0;ll */ LL dfs(int pos ,int x,int limit ,int status) { if(pos<0) return status; if(dp[pos][x][status]!=-1&&!limit) return dp[pos][x][status]; int endi=9; if(limit) endi = num[pos]; LL res = 0; for(int i=0;i<=endi;i++) { res+=dfs(pos-1 ,i ,limit&&(i==endi) ,status||(i==9&&x==4) ); } if(!limit) dp[pos][x][status] = res; return res; } int main() { memset(dp,-1,sizeof(dp)); int _; scanf("%d",&_); while(_--) { LL n; scanf("%I64d",&n); LL tem = n; len = 0; while(tem) { num[len++]=tem%10; tem/=10; } LL m = dfs(len-1,0,1,0); printf("%I64d\n",m); } return 0; }
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