【 CodeForces 612A 】 The Text Splitting

The Text Splitting

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status
Description
You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string “Hello” for p = 2, q = 3 can be split to the two strings “Hel” and “lo” or to the two strings “He” and “llo”.

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input
The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output
If it’s impossible to split the string s to the strings of length p and q print the only number “-1”.

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Sample Input
Input
5 2 3
Hello
Output
2
He
llo
Input
10 9 5
Codeforces
Output
2
Codef
orces
Input
6 4 5
Privet
Output
-1
Input
8 1 1
abacabac
Output
8
a
b
a
c
a
b
a
c
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因为一开始想错了，所以印象特别深的一题，后来还看了别人的代码⊙﹏⊙。一开始以为是有三种情况 p 的倍数，q的倍数，p+q的倍数（如14 =（3+4）*2），后来发现WA 了，提示Wrong answer on test 9。

后来发现自己错的地方不是细节，而是思路，因为还有 n倍的（p的倍数+q的倍数）这种情况没有考虑（如 22 =（5+3+3）*2）。思路错了，如果一开始想到是这样的话就会直接用两个for，结果费了好大力气。

下面贴上错误和正确的两个代码：
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错误代码：

//三个搜索……
#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;


int main()
{
    int n,p,q,i,j,a,b,c,t,tt,ttt,k;
    char ch[105];
    while(~scanf("%d%d%d",&n,&p,&q))
    {
        k=t=tt=ttt=0;
        getchar();
        for( i=0; i<n; i++ )
        {
            ch[i]=0;
        }
        for( i=0; i<n; i++ )
        {
            scanf("%c",&ch[i]);
        }
        a=n%p;
        b=n%q;
        c=n%(p+q);
        if(!a)
        {
            t=n/p;
            printf("%d\n",t);
            k=0;
            for( j=0; j<t; j++ )
            {   
                for( i=0; i<p; i++ )
                {
                    printf("%c",ch[i+p*k]); 
                }
                    printf("\n"); 
                    k++;
            }
            continue;
        }
        if(!b)
        {
            tt=n/q;
            printf("%d\n",tt);
            k=0;
            for( j=0; j<tt; j++ )
            {   
                for( i=0; i<q; i++ )
                {
                    printf("%c",ch[i+q*k]); 
                }
                    printf("\n"); 
                    k++;
            }
            continue;
        }
        if(!c)
        {
            ttt=(n/(p+q))*2;
            printf("%d\n",ttt);
            k=0;
            for( j=0; j<n/(p+q); j++ )
            {   
                for( i=0; i<p; i++ )
                {

                    printf("%c",ch[i+(p+q)*k]); 
                }
                printf("\n"); 
                for( i=p; i<q+p; i++)
                {
                    printf("%c",ch[i+(p+q)*k]);
                }
                printf("\n"); 
                k++;
            }
            continue;
        }

        if( a != 0 && b != 0 && c != 0 )
        {
            printf("-1\n");
            continue;
        }
    }
    return 0;       
}

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正确代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
char s[110];

int main()
{
    int n,p,q,count,i,j,k;
    while( ~scanf("%d%d%d%s",&n,&p,&q,s) )
    {
        count=-1;
        for( i=0; i<=n/p; i++ )
        {   
            for( j=0; j<=n/q; j++ )
            if( i*p + j*q == n )    //把每一种可能都试一遍 
            {
                count=i+j;          // 前面错误的原因是因为没有考虑多个q 和单个 p，或者单个 p 和多个q
                                      // 如果题意没有理解错，就会想到用两个循环，而不是分开来写 
                printf("%d\n",count); 
                    for( k=0; k<n; k++ )
                    {
                        printf("%c",s[k]);
                        if( i && k<i*p && !((k+1)%p) || j && k >= i*p &&!( (k-i*p+1)%q) )
                        printf("\n");   //确定换行的时间 
                    } 
                break;
            }
            if( count != -1 )break;
        }
        if( count == -1 )printf("-1\n");
    }
    return 0;
}