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;
- }
- }
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