Codeforces 598B Queries on a String 【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 li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki 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 li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 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.

题意：给定一个字符串s，下标从1 - |s|。

有m次操作，L R k - 表示讲[L, R]里面的字符右移k次，第R位右移一次到达第L位。问你经过m次变化后的字符串。

思路：优化区间长度k % (R-L+1)。先从后向前扫一遍，求出当前位置i经过k次变化后到达的位置。最后从前向后扫一遍，更新字符即可。时间复杂度O(m|s|)。

AC代码:

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-4#define MAXN (10000+10)#define MAXM (1000000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 100000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;char str[MAXN], Cstr[MAXN];bool vis[MAXN];int main(){    Rs(str);    int m; Ri(m);    while(m--)    {        int l, r, k;        Ri(l); Ri(r); Ri(k);        int len = (r - l + 1);        k %= len;        CLR(vis, false);        for(int i = r-1; i >= l-1; i--)        {            int pos = (i - l + 1 + k) % len + l-1;            Cstr[pos] = str[i];        }        for(int i = l-1; i <= r-1; i++)            str[i] = Cstr[i];    }    Ps(str);    return 0;}