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ZOJ1042

思路：把三種字符分別存儲，在+k1、+k2、+k3後取模

1.要考慮到k1>l1等情況
2.分母不能爲0，否則會出現 Floating Point Error

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
char str[1000000];
int main()
{
    int k1,k2,k3;
    int l1,l2,l3;
    char s1[5000],s2[5000],s3[5000];
    while(cin>>k1>>k2>>k3&&(k1+k2+k3))
    {
        cin>>str;
        l1=l2=l3=0;
        int len=strlen(str),i;
        for(i=0;i<len;i++)
        {
            if(str[i]>='a'&&str[i]<='i')s1[l1++]=str[i];
            else if(str[i]>='j'&&str[i]<='r')s2[l2++]=str[i];
            else s3[l3++]=str[i];
        }
       // for(i=0;i<l1;i++)printf("*%c\n",s1[i]);
        int p1,p2,p3;

        //1.要考慮到k1>l1等情況
        //2.分母不能爲0，否則會出現 Floating Point Error
        if(k1>l1&&l1!=0)k1=k1%l1; 
        if(k2>l2&&l2!=0)k2=k2%l2;
        if(k3>l3&&l3!=0)k3=k3%l3;

        p1=l1-k1;
        p2=l2-k2;
        p3=l3-k3;
        for(i=0;i<len;i++)
        {
            if(str[i]>='a'&&str[i]<='i'){
                    str[i]=s1[(p1++)%l1];
          //          cout<<str[i]<<endl;
            }
            else if(str[i]>='j'&&str[i]<='r')str[i]=s2[(p2++)%l2];
            else str[i]=s3[(p3++)%l3];
        }
        cout<<str<<endl;

    }
    return 0;
}