HDU2772 Matchsticks 【模拟题】

Matchsticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 804    Accepted Submission(s): 282

Problem Description

Matchsticks are ideal tools to represent numbers. A common way to represent the ten decimal digits with matchsticks is the following:
This is identical to how numbers are displayed on an ordinary alarm clock. With a given number of matchsticks you can generate a wide range of numbers. We are wondering what the smallest and largest numbers are that can be created by using all your matchsticks.

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:
* One line with an integer n (2 ≤ n ≤ 100): the number of matchsticks you have.

Output

Per testcase:
* One line with the smallest and largest numbers you can create, separated by a single space. Both numbers should be positive and contain no leading zeroes.

Sample Input

436715

Sample Output

7 76 1118 711108 7111111

题意：n个火柴棒所能摆出的最小数和最大数 

n=2  1        1
n=3  7        7
n=4  4        11
n=5  2         71
n=6  6         111
n=7  8         711
n=8  10       1111
n=9  18       7111
n=10 22      11111
n=11 20       71111
n=12 28       111111
n=13 68       711111
n=14 88       1111111
n=15 108     7111111
n=16 188     11111111
n=17 200     71111111
n=18 208     111111111
n=19 288     711111111
n=20 688     1111111111
n=21 888     7111111111
n=22 1088   11111111111
n=23 1888   71111111111
n=24 2008   111111111111
n=25 2088   711111111111
n=26 2888   1111111111111
n=27 6888   7111111111111
n=28 8888   11111111111111
n=29 10888 71111111111111

找规律,

最小数 从 n [ 15 -> 21 ] 开始 周期 T = 7 每加一个周期多一个 8

最大数 从 n [ 2 -> 3 ]      开始 周期 T = 2 每加一个周期多一个 1

#include<stdio.h>int ans[22][2]={{0},{0},{1,1},{7,7},{4,11},{2},{6},{8},{10},{18},{22},{20},{28},{68},{88},{108},{188},{200},{208},{288},{688},{888}};int main(){int t;int i,j;int num;int n8,n1;scanf("%d",&t);while(t--){scanf("%d",&num);if(num<=21){printf("%d ",ans[num][0]);n1=(num-2)/2;num%2?printf("7"):printf("1");while(n1--){printf("1");}}else{printf("%d",ans[(num-15)%7+15][0]);n8=(num-15)/7;while(n8--){printf("8");}n1=(num-2)/2;num%2?printf(" 7"):printf(" 1");while(n1--){printf("1");}}printf("\n");}return 0;}