POJ1107 W's Cipher 简单模拟

POJ1107 W’s Cipher

Description
先是将字母和下划线分为三类a~i，j~r，s~z和‘_’; 然后，独立地交换每组的字母，例如：对于 “_icuo_bfnwhoq_kxert”，第一组符号有{i,c,b,f,h,e}，位置分别为{2,3,7,8,11,17}，假设k1=2；那么第一组符号顺序变为{h,e,i,c,b,f}，不过在原字符串中占有的位置还是{2,3,7,8,11,17}，然后第二组······第三组······

Sample Input
2 3 1
_icuo_bfnwhoq_kxert
1 1 1
bcalmkyzx
3 7 4
wcb_mxfep_dorul_eov_qtkrhe_ozany_dgtoh_u_eji
2 4 3
cjvdksaltbmu
0 0 0

Sample Output
the_quick_brown_fox
abcklmxyz
the_quick_brown_fox_jumped_over_the_lazy_dog
ajsbktcludmv

题解：
整理ing

AC_Code(cpp)：

#include<iostream>using namespace std;struct mychar{    char character;    int position;};void Decrypted(char encrypted[], int k1, int k2, int k3){    char decrypted[100]="";    mychar char1[100];    mychar char2[100];    mychar char3[100];    int count1 = 0;    int count2 = 0;    int count3 = 0;    int len = strlen(encrypted);    for (int i = 0; i < len; i++)    {        if (encrypted[i] >= 's' || encrypted[i] == '_')        {            char3[count3].character = encrypted[i];            char3[count3].position = i;            count3++;        }        else if (encrypted[i] >= 'j')        {            char2[count2].character = encrypted[i];            char2[count2].position = i;            count2++;        }        else        {            char1[count1].character = encrypted[i];            char1[count1].position = i;            count1++;        }    }    for (int i = 0; i < count3; i++)    {        decrypted[char3[(i + k3) % count3].position] = char3[i].character;    }    for (int i = 0; i < count2; i++)    {        decrypted[char2[(i + k2) % count2].position] = char2[i].character;    }    for (int i = 0; i < count1; i++)    {        decrypted[char1[(i + k1) % count1].position] = char1[i].character;    }    cout << decrypted << endl;}int main(){    int k1, k2, k3;    char encrypted[100]="";    while (cin >> k1 >> k2 >> k3 && (k1 + k2 + k3))    {        cin >> encrypted;        Decrypted(encrypted, k1, k2, k3);    }    return 0;}