poj 1107W's Cipher

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题意:

给定字符串可分为三个部分:Encrypting a message requires three integer keys, k1, k2, and k3. The letters [a-i] form one group, [j-r] a second group, and everything else ([s-z] and underscore) the third group.k1代表first group 右移的步数（仅在这一部分中），k2, k3同理。

思路:很简单，对每一部分进行移位。可另设数组存储每一部分，对原串进行更改。

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define FOR(i, n) for(i = 0; i < (n); ++i)#define FORE(i, a, b) for(i = (a); i <= (b); ++i)struct node{    char c;    int id;};int main(){    int k1, k2, k3, i, len;    char s[100];    node g1[100],g2[100],g3[100];    while(scanf("%d %d %d", &k1, &k2, &k3)!=EOF && (k1+k2+k3)!=0){        memset(s, 0, sizeof(s));        scanf("%s", s);        len = strlen(s);        int c1 = 0, c2 = 0, c3 = 0;        FOR(i, len){            if(s[i] >= 'a' && s[i] <= 'i'){                g1[c1].id = i;                g1[c1++].c = s[i];            }            else if(s[i] >= 'j' && s[i] <= 'r'){                g2[c2].id = i;                g2[c2++].c = s[i];            }            else{                g3[c3].id = i;                g3[c3++].c = s[i];            }        }        FOR(i, c1)            s[g1[(i+k1)%c1].id] = g1[i].c;        FOR(i, c2)            s[g2[(i+k2)%c2].id] = g2[i].c;        FOR(i, c3)            s[g3[(i+k3)%c3].id] = g3[i].c;        printf("%s\n",s);    }    return 0;}