模拟就好了 坑比较多
1.当‘-’两端的字符分别是数字和字母时，原样输出。
2.字符串首尾的‘-’需要原样输出。
3.连续的‘-’需要原样输出。
4.存放展开字符串的数组开大点
附几组样例：
输入：
2 4 2
a-b-c
输出：
abc
输入：
2 2 2
9-a-a
输出：
9-a-a
输入：
1 5 1
-254-243-52-345-243-5234-52-345-234-52-345-234-52345-4325-2345-2345-2345
输出：
-254-2434444452345-24344444523452345-23452345-23452345-4325-2345-2345-2345
输入：
1 1 1
2-43-sdf-ewr0-j-9re-j0g-9e0-9as-d09jf-9asdjf0q-w-ejr0q-59jdsnf-9z-x9v0-sd9fn--q
输出：
2343-sdf-ewr0-j-9refghij0g-9e0123456789as-d09jf-9asdjf0qrstuvw-ejr0q-59jdsnf-9z-x9v0-sd9fn--q
输入：
2 8 2
--09-8-w-er-7h-08w-e7-hc-r890-q7w-eh-rc98-07-q8-ewr-8h-c-8-294-5-dsf--k-h-2-48-3k-h-sd-fq-a-
输出：
--09-8-w-er-7h-08w-e7-hcQQQQQQQQPPPPPPPPOOOOOOOONNNNNNNNMMMMMMMMLLLLLLLLKKKKKKKKJJJJJJJJIIIIIIIIHHHHHHHHGGGGGGGGFFFFFFFFEEEEEEEEDDDDDDDDr890-q7w-ehQQQQQQQQPPPPPPPPOOOOOOOONNNNNNNNMMMMMMMMLLLLLLLLKKKKKKKKJJJJJJJJIIIIIIIIrc98-07-q8-ewr-8h-c-8-2945-dsf--k-h-23333333348-3k-hRRRRRRRRQQQQQQQQPPPPPPPPOOOOOOOONNNNNNNNMMMMMMMMLLLLLLLLKKKKKKKKJJJJJJJJIIIIIIIIsdEEEEEEEEfq-a-
绝了 恶心
//P1098 [NOIP2007 提高组] 字符串的展开
//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>//abs
#include <cctype>//isdigit
#define clr(x) memset(x,0,sizeof(x))
using namespace std;
char str[105],buf[10500];
int len,idx;
void strUnfold(int p,int p2,int p3) {
idx = 0;
int ptr = 0,step;
for(int i = 0;i<len;i++) {
if(str[i]!='-') {
buf[idx++] = str[i];
ptr = idx;
continue;
}
else {
if((i==len-1&&str[i]=='-')||(i==0&&str[i]=='-')||(str[i+1]=='-'||str[i-1]=='-')||(isdigit(str[i-1])&&isalpha(str[i+1]))||(isdigit(str[i+1])&&isalpha(str[i-1]))||(str[i-1]>=str[i+1])) {
buf[idx++] = '-';
continue;
}
step = str[i+1]-str[i-1]-1;
int k = p2;
if(p==3) {
for(int j = 1;j<=step*k;j++) {
buf[idx++] = '*';
}
}
else if(isdigit(str[i-1])&&p!=3) {
int flag = 0;
for(int j = 1;j<=step;j++) {
for(int m = 1;m<=k;m++) {
buf[idx++] = str[i-1]-'0'+j+48;
}
flag = 1;
}
if(p3==2&&flag) {
//逆序
int t = 1;
for(int q = ptr;q<(idx+ptr)/2;q++) {
char tmp;
tmp = buf[q];
buf[q] = buf[idx-t];
buf[idx-t] = tmp;
t++;
}
}
}
else if(isalpha(str[i-1])) {
int flag = 0;
if(p==1) {
for(int j = 1;j<=step;j++) {
for(int m = 1;m<=k;m++) {
buf[idx++] = str[i-1]-'a'+j+97;
}
flag = 1;
}
if(p3==2&&flag) {
//逆序
int t = 1;
for(int q = ptr;q<(idx+ptr)/2;q++) {
char tmp;
tmp = buf[q];
buf[q] = buf[idx-t];
buf[idx-t] = tmp;
t++;
}
}
}
else if(p==2) {
for(int j = 1;j<=step;j++) {
for(int m = 1;m<=k;m++) {
buf[idx++] = str[i-1]-'a'+j+97-32;
}
flag = 1;
}
if(p3==2&&flag) {
//逆序
int t = 1;
for(int q = ptr;q<(idx+ptr)/2;q++) {
char tmp;
tmp = buf[q];
buf[q] = buf[idx-t];
buf[idx-t] = tmp;
t++;
}
}
}
}
}
}
}
int main() {
#ifdef LOCAL
freopen("data.in","r",stdin);
freopen("data.out","w",stdout);
#endif
int p1,p2,p3;
cin>>p1>>p2>>p3;
scanf("%s",str);
len = strlen(str);
strUnfold(p1,p2,p3);
for(int i = 0;i<idx;i++) {
cout<<buf[i];
}
}