Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2123 Accepted Submission(s): 1019
题解:没写出来,看了人家的修改了自己的,然而wawawa;
我的:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
typedef long long LL;
const int MAXN=250010;
char s[MAXN];
int k;
int len;
//int i;
int dfs(int x){
int temp,e;
for(int i=x;i<len;i++){
//temp=1;
if(s[i]==‘)‘)return i+1;
/*if(s[i]>=‘0‘&&s[i]<=‘9‘){
temp=s[i]-‘0‘;
i++;
while(s[i]>=‘0‘&&s[i]<=‘9‘){
temp=temp*10+s[i]-‘0‘;i++;
}
}*/
for(temp=0;isdigit(s[i]);i++)
temp=temp*10+s[i]-‘0‘;
if(!temp) temp=1;
if(s[i]==‘(‘){
while(temp--){
e=dfs(i+1);
}
i=e;
}
else{
while(temp--)printf("%c",s[i]);
}
}
}
int main(){
int T;
SI(T);
T_T{
memset(s,0,sizeof(s));
scanf("%s",s);
len=strlen(s);
dfs(0);puts("");
}
return 0;
}
大神ac;
#include <iostream>
#include <cctype>
#include <cstring>
#include <string>
using namespace std;
string s;
int fun(int ith)
{
int k,e;
char c;
for(c=s[ith++];ith<s.size()&&c!=‘)‘;c=s[ith++])//递归结束的条件是字符串结束或遇到右括号
{
for(k=0;isdigit(c);c=s[ith++])
k=k*10+c-‘0‘;
if(!k) k=1;
if(c==‘(‘){
while(k--)
e=fun(ith);
ith=e;//重置ith的值,到下层递归结束的位置
}
else
{
while(k--)
putchar(c);
}
}
if(c==‘)‘) return ith;//返回本次读到结尾的位置
}
int main()
{
int i,j,k,T;
cin>>T;
while(T--)
{
s.clear();
cin>>s;
fun(0);//进入递归
cout<<endl;
}
return 0;
}
原文:http://www.cnblogs.com/handsomecui/p/5087251.html