• 1031. Hello World for U (20)


    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:

    h  d
    e  l
    l  r
    lowo
    

    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.

    Input Specification:

    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.

    Output Specification:

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

    Sample Input:

    helloworld!
    

    Sample Output:

    h   !
    e   d
    l   l
    lowor

    题目大意:把字符串的输出按U形样式输出。
    解题思路:可根据公式n1=n3=max{k|k<=n2 for all 3<=n2<=N}和n1+n2+n3-2=N,对n1与n2进行遍历,先求解出n1和n2的值。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int main(){
    	char s1[81],s2[81][81];
    	scanf("%s",s1);
    	int length=strlen(s1);
    	int i,j;
    	int n = length+2;
    	int flag=0;
    	for(i=1;i<=length;i++){
    		for(j=1;j<=i;j++){
    			if((i+j*2)%n == 0){
    				flag = 1;
    				break;
    			}
    		}
    		if(flag==1)break;
    	}
    	int n2=i-2;
    	int n3,n1;
    	n3=n1=j;
    	length-=1;
    	for(i=0;i<n1-1;i++){
    		printf("%c",s1[i]);
    		for(j=0;j<n2;j++){
    			printf(" ");
    		}
    		printf("%c
    ",s1[length-i]);
    	}
    	for(i=n1-1;i<=length-n1+1;i++){
    		printf("%c",s1[i]);
    	}
    	printf("
    ");
    	return 0;
    } 
    

      

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  • 原文地址:https://www.cnblogs.com/grglym/p/7697740.html
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