2015 Multi-University Training Contest 2 1009 I Wanna Become A 24-Point Master(HDU5308) 构造-优快云博客

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I Wanna Become A 24-Point Master

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 757 Accepted Submission(s): 315
Special Judge

Problem Description

Recently Rikka falls in love with an old but interesting game -- 24 points. She wants to become a master of this game, so she asks Yuta to give her some problems to practice.

Quickly, Rikka solved almost all of the problems but the remained one is really difficult:

In this problem, you need to write a program which can get 24 points with

n numbers, which are all equal to

n .

Input

There are no more then 100 testcases and there are no more then 5 testcases with

n≥100 . Each testcase contains only one integer

n (1≤n≤105)

Output

For each testcase:

If there is not any way to get 24 points, print a single line with -1.

Otherwise, let

A be an array with

2n−1 numbers and at firsrt

Ai=n (1≤i≤n) . You need to print

n−1 lines and the

i th line contains one integer

a , one char

b and then one integer c, where

1≤a,c<n+i and

b is "+","-","*" or "/". This line means that you let

Aa and

Ac do the operation

b and store the answer into

An+i .

If your answer satisfies the following rule, we think your answer is right:

1.

A2n−1=24

2. Each position of the array

A is used at most one tine.

3. The absolute value of the numerator and denominator of each element in array

A is no more than

109

Sample Input

Sample Output

Source

2015 Multi-University Training Contest 2

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题意：

1个1，2个2，3个3，4个4,5个5…… 100000个100000用加减乘除括号算24。

输出的格式是，一个数组有2 * n - 1个数，前n个数都为n，则输出n - 1个式子，含义是下标为x的与下标y的做ＸＸ操作，结果会被保存在下一个未保存数字的单元内，要让第2 * n - 1个数为24

思路：

用12个n可以用如下的方法算24：

(n + n + n + n) / n * (n + n + n + n + n + n) / n = 24

跟n的值无关，所以只要不小于12的偶数都可以用此方法，先用前12个算出24，然后+n - n 就行了

同理，13个n可以用如下的方法算24，这样可以解决不小于13的奇数：

(n + n + n ) / n * (n + n + n + n + n + n + n + n) / n = 24

这样12及以后的就都搞定了，12之前的手算一下就可以打表了。

#include <cstdio>

using namespace std;
///1——11的表
char str[12][150]= {"",
                    "-1",
                    "-1",
                    "-1",
                    "1 * 2\n5 + 3\n6 + 4",
                    "1 * 2\n3 * 6\n7 - 4\n8 / 5",
                    "1 + 2\n3 + 7\n4 + 8\n5 + 9\n10 - 6",
                    "1 + 2\n3 + 8\n4 + 5\n6 + 10\n11 / 7\n9 + 12",
                    "1 + 2\n3 + 9\n4 - 5\n11 * 6\n12 * 7\n13 * 8\n14 + 10",
                    "1 + 2\n10 + 3\n11 / 4\n5 * 12\n6 + 7\n14 + 8\n15 / 9\n13 - 16",
                    "1 + 2\n3 + 11\n4 + 5\n13 + 6\n14 + 7\n15 + 8\n16 + 9\n17 / 10\n12 - 18",
                    "1 + 2\n3 + 4\n13 / 5\n12 + 14\n6 - 7\n8 - 9\n10 - 11\n15 + 16\n18 + 17\n19 + 20",
                   };

int main() {
    int n;
    while(~scanf("%d",&n)) {
		///1——11的按表输出
        if(n < 12) {
            puts(str[n]);
		///超过11，经行构造，前面一截是一样的
        } else {
            printf("1 + 2\n3 + 4\n5 + 6\n7 + 8\n9 + 10\n");
			///偶数的构造方法 （4*6）
            if(n % 2 == 0) {
                printf("%d + %d\n",n + 1,n + 2);
                printf("%d + %d\n",n + 3,n + 4);
                printf("%d + %d\n",n + 5,n + 7);
                printf("%d / %d\n",n + 6,11);
                printf("%d / %d\n",n + 8,12);
                printf("%d * %d\n",n + 9,n + 10);
				///然后补+n-n
                for(int i = 13; i < n; i += 2) {
                    printf("%d + %d\n%d - %d\n",n + i - 2,i,n + i - 1,i + 1);
                }
			///奇数的构造方法（3*8）
            } else {
                printf("%d + %d\n",n + 1,11);
                printf("%d + %d\n",n + 2,n + 3);
                printf("%d + %d\n",n + 4,n + 5);
                printf("%d + %d\n",n + 7,n + 8);
                printf("%d / %d\n",n + 6,12);
                printf("%d / %d\n",n + 9,13);
                printf("%d * %d\n",n + 10,n + 11);
				///然后补+n-n
                for(int i = 14;i < n;i += 2){
                    printf("%d + %d\n%d - %d\n",n + i - 2,i,n + i - 1,i + 1);
                }
            }
        }
    }
    return 0;
}