【Educational Codeforces Round 1B】【字符串平移 水题】Queries on a String 字符串平移水题

B. Queries on a String

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string s and should process m queries. Each query is described by two 1-based indices l_i, r_i and integer k_i. It means that you should cyclically shift the substring s[l_i... r_i] k_i times. The queries should be processed one after another in the order they are given.

One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.

For example, if the string s is abacaba and the query is l₁ = 3, r₁ = 6, k₁ = 1 then the answer is abbacaa. If after that we would process the query l₂ = 1, r₂ = 4, k₂ = 2 then we would get the string baabcaa.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.

Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.

The i-th of the next m lines contains three integers l_i, r_i and k_i (1 ≤ l_i ≤ r_i ≤ |s|, 1 ≤ k_i ≤ 1 000 000) — the description of the i-th query.

Output

Print the resulting string s after processing all m queries.

Sample test(s)

input

abacaba23 6 11 4 2

output

baabcaa

Note

The sample is described in problem statement.

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=10010,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;char s[N];char tmp[N];int main(){while(~scanf("%s",s+1)){int m;scanf("%d",&m);int l,r,k;while(m--){scanf("%d%d%d",&l,&r,&k);k%=(r-l+1);int g=0;for(int i=r-(k-1);i<=r;i++)tmp[++g]=s[i];for(int i=l;i<=r-k;i++)tmp[++g]=s[i];for(int i=1,j=l;i<=g;i++,j++)s[j]=tmp[i];}puts(s+1);}return 0;}/*【题意】给你一个长度在[1,10000]范围内的字符串，让你执行m次操作对于每个操作，我们要对这个字符串[l,r]区间范围的数，向右做旋转式平移k次（注，向右的旋转式平移就是位置p是由位置p-1的数转移得，位置l是由位置r转移得）让你输出所有操作完成之后的串。（1<=m<=300，1<=l<=r<=|s|，1<=k<=1e6）【类型】字符串操作水题【分析】首先k很大，但是显然可以使得k%=(r-l+1)，提高运行效率。然后我们把右移溢出的k个字符先提取过来——for(int i=r-(k-1);i<=r;i++)tmp[++g]=s[i];然后再把右移无溢出的剩余字符提取过来——for(int i=l;i<=r-k;i++)tmp[++g]=s[i];接着还原即可——for(int i=1,j=l;i<=g;i++,j++)s[j]=tmp[i];【时间复杂度&&优化】O(m*|s|)*/