TOJ 3850: String Function Encoding

传送门：http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=3850

时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte

描述

Bessie discovered a new function that the entire herd can apply to its character strings.
Given both a number N (1 <= N <= 15) and a
string S, with length strictly greater than N, define f(N, S) as a new string
composed
of the concatenation of the substring from character N (zero based
-- first character is number 0) through the end of S and the string S
itself.
For example, with N = 2, and S = "COW", f(N, S) = "W" + "COW" =
"WCOW". Also, f(3, "USACO") = "CO" + "USACO" = "COUSACO".
Bessie is
enthralled with this function and wants to iterate it several times. For
example, if she iterates the function once for
"COW" and N = 2, she will get
"WCOW". If she applies the function with N = 2 again to that string, she will
get "OWWCOW", and if she applies it one more time with N = 2, she will get
"WCOWOWWCOW".
Help Bessie encode a total of Z (1 <= Z <= 100)
strings, str_1, str_2, and so on.  Each str_i has length in the range 2..100
and
contains only upper case letters. Each string is presented with its own
N_i (0 <= N_i < length(str_i), and iteration count C_i (1 <= C_i <=
12).

输入

* Line 1: A single integer: Z
* Lines 2..Z+1: Line i+1 contains two
space-separated integers, a space, and string to be encoded: N_i, C_i, and
str_i

输出

* Lines 1..Q: Line j contains the iterated, encoded version of
str_j

样例输入

2
2 3 COW
3 2 USACO

样例输出

WCOWOWWCOW
SACOCOUSACO

提示

OUTPUT DETAILS:
The arrow denotes an iteration of the function

COW -> WCOW -> OWWCOW -> WCOWOWWCOW

USACO -> COUSACO -> SACOCOUSACO

题意：就是把得到的字符串的 从N开始到串结束 这个子串，截取来，放到该串的最前面，这算一次操作

比如说第一组数据，N = 2 ,F =3 代表要执行3次，每次把  第二个位置到最后一位 的子串，截取下来，放到最前面。

思路：拿string里面的substr和insert ，截取和插入一下就好了，水题！

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
　　 int m,k,T,n;
    scanf("%d",&T);
    while(T--)
    {
        string s;
        scanf("%d %d",&n,&m);
        cin>>s;
        int t = n;
        while(m--)
        {
            string s1 = s.substr(t,s.size()-t+);
            s.insert(,s1);
        }
        cout<<s<<endl;
    }
}