HDU 1297:Children’s Queue
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Children’s Queue
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 8 Accepted Submission(s) : 1
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Problem Description
There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
Input
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)
Output
For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.
Sample Input
123
Sample Output
124
Author
Source
杭电ACM集训队训练赛(VIII)
你 离 开 了 , 我 的 世 界 里 只 剩 下 雨 。 。 。
import java.util.*;import java.math.*;public class Main{ public static void main(String[] args){ Scanner cin = new Scanner(System.in); BigInteger []num = new BigInteger[1010]; num[1] = BigInteger.valueOf(1); num[2] = BigInteger.valueOf(2); num[3] = BigInteger.valueOf(4); num[4] = BigInteger.valueOf(7); for(int i=5; i<1001; i++) num[i] = num[i-1].add(num[i-2].add(num[i-4])); int n; while(cin.hasNext()){ n = cin.nextInt(); System.out.println(num[n]); } }}
本来想用C/C++做这道题,结果看到(1<=n<=1000),想到后面估计数据规模非常大,用“__Int64”估计也不行,所以想到了java里面的BigInteger类型,因此就有了上面的代码咯!!
下面这个C/C++语言AC的代码是转载的……
#include<stdio.h>int f[1001][101];int main(){ int n; f[0][1] = 1; f[1][1] = 1; f[2][1] = 2; f[3][1] = 4; for(int i = 4; i < 1001; ++i) for(int j = 1; j < 101; ++j) { f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j]; f[i][j + 1] += f[i][j] / 10000; f[i][j] %= 10000; } while(scanf("%d", &n) != EOF) { int k = 100; while(!f[n][k--]); printf("%d", f[n][k + 1]); for(; k > 0; --k) printf("%04d", f[n][k]); printf("\n"); } return 0;}
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