题意简述:
给出一个字符串,每次给定l,r,k,选择一个子串l-r,然后子串向右移动k个单位.
做法:
每次k对子串的长度取模,然后模拟即可(使用substr函数截取前半段和后半段,交换前半段和后半段即可)
点击查看代码
// Problem: B. Queries on a String
// Contest: Codeforces - Educational Codeforces Round 1
// URL: https://codeforces.com/contest/598/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: a2954898606
// Create Time:2023-11-29 13:32:43
//
// Powered by CP Editor (https://cpeditor.org)
/*
/\_/\
(o o)
/ \
// ^ ^ \\
/// \\\
\ |
\_____/
/\_ _/\
*/
#include<bits/stdc++.h>
#define all(X) (X).begin(),(X).end()
#define all2(X) (X).begin()+1,(X).end()
#define pb push_back
#define mp make_pair
#define sz(X) (int)(X).size()
#define YES cout<<"YES"<<endl
#define Yes cout<<"Yes"<<endl
#define NO cout<<"NO"<<endl
#define No cout<<"No"<<endl
using namespace std;
typedef long long unt;
typedef vector<long long> vt_unt;
typedef vector<int> vt_int;
typedef vector<char> vt_char;
template<typename T1,typename T2> inline bool ckmin(T1 &a,T2 b){
if(a>b){
a=b;
return true;
}
return false;
}
template<typename T1,typename T2> inline bool ckmax(T1 &a,T2 b){
if(a<b){
a=b;
return true;
}
return false;
}
namespace smart_math{
long long quickpow(long long a,long long b,long long mod=0){
long long ret=1;
while(b){
if(b&1)ret*=a;
if(mod)ret%=mod;
b>>=1;
a*=a;
if(mod)a%=mod;
}
if(mod)ret%=mod;
return ret;
}
long long quickmul(long long a,long long b,long long mod=0){
long long ret=0;
while(b){
if(b&1)ret+=a;
if(mod)ret%=mod;
b>>=1;
a=a+a;
if(mod)a%=mod;
}
if(mod)ret%=mod;
return ret;
}
long long ceil_x(long long a,long long b){
if(a%b==0)return a/b;
else return (a/b)+1;
}
}
using namespace smart_math;
const long long mod=1e9+7;
const long long N=310000;
const long long M=300;
int solve(){
string s;
cin>>s;
s=" "+s;
int q;
cin>>q;
while(q--){
int l,r,k;
cin>>l>>r>>k;
int len=r-l+1;
k%=len;
int m=len-k;//front
//cout<<"m:"<<m<<endl;
string t=s.substr(l,m);
//cout<<"t:"<<t<<endl;
int cnt=0;
for(int i=r-k+1;i<=r;i++){
s[l+cnt]=s[i];
cnt++;
}
cnt=0;
for(int i=l+k;i<=r;i++){
s[i]=t[cnt];
cnt++;
}
//cout<<s<<endl;
}
for(int i=1;i<sz(s);i++){
cout<<s[i];
}
cout<<endl;
return 0;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
t=1;
//cin>>t;
while(t--){
int flag=solve();
if(flag==1) YES;
if(flag==-1) NO;
}
return 0;
}