PAT甲级.1031. Hello World for U (20)

1031. Hello World for U (20)

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题目：

Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, “helloworld” can be printed as:

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible – that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.

输入格式：

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

输出格式：

For each test case, print the input string in the shape of U as specified in the description.

输入样例：

helloworld!

输出样例：

PAT链接

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题意：

输入一个字符串长的N，将该字符串按U型输出。有n1+n2+n3-2=N

n1 == n3即左侧字符数等于右侧图案数n2>=n1 即底部图案数不小于两边的图案数满足以上条件情况下使n1尽可能大

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思路：

输入得到字符串长度： gets(str); int len = strlen(str); 分别计算得到n1,n2,n3 int H = (len+2)/3-1; //n1-1 int bottom = (len+2)/3; //n2(下面还会修正） …… if(remainder == 1) bottom++; //n2修正 else if(remainder == 2) bottom+=2; //n2修正 else ; 先输出n1和n3,最后输出n2(也可考虑通过二维数组的方式输出)

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代码：

/** * @tag PAT_A_1031 * @authors R11happy (xushuai100@126.com) * @date 2016-8-15 15:42-16:02 * @version 1.0 * @Language C++ * @Ranking 220/3486 * @function null */ #include <cstdio> #include <cstdlib> #include <cstring> int main(int argc, char const *argv[]) { char str[100]; gets(str); int len = strlen(str); int H = (len+2)/3-1; //n1-1 int bottom = (len+2)/3; //n2 int remainder = (len+2)%3; if(remainder == 1) bottom++; //n2修正 else if(remainder == 2) bottom+=2; //n2修正 else ; int i; for(i=0; i<H; i++) { printf("%c",str[i] ); //打印左侧字符 for(int j=0; j<(bottom-2); j++) { printf(" "); } printf("%c\n",str[len-i-1] ); //打印右侧字符 } //打印底部字符 for(int k=0; k<bottom; k++,i++ ) { printf("%c",str[i] ); } printf("\n"); return 0; }

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收获：

1.gets函数在vs2015不能用，要改成gets_s，不过gets_s在PAT中编译出错 2.数学推导后再输出，其中求n1,n2,n3也可以用如下代码：

int n1 = (N+2) / 3; int n3 = n1; int n2 = N+2-n1-n3;