【牛客比赛】小Q喜欢撸串（字符串）
链接：https://ac.nowcoder.com/acm/problem/23905
来源：牛客网
description
小Q挺喜欢撸串的，没错，字符串！
你给小Q送上了n个字符串
对于一个字符串s，如果在小Q撸掉(删除)任意个字符之后，"NowCoder"是其子串，则这个字符串s是可撸的。小Q最近切题切到手软，想撸串散散心。如果你给他呈现的字符串是可撸的，他会很开心，否则他会很桑心。
输入描述:
一个整数n，表示字符串的数量
接下来每行一个字符串sis_isi，表示小Q看到的第i个字符串
输出描述:
输出有n行，如果小Q开心他会说QAK，否则他会说QIE
示例1
输入
2
NowCoder
NoowCoder
输出
QAK
QAK
说明
第一个字符串不需要撸掉任何字符
第二个字符串撸掉第二个或者第三个都可以
示例2
输入
5
Nowcoder
nowcoder
NowCoder
NowoowCoder
NOI2019NowCoder
输出
QIE
QIE
QAK
QAK
QAK
分析：
快读，读入字符和整型数字，利用getchar进行读入，因getchar比任何读入方式都快
尽量使用快读或者scanf，以防卡时间
inline int read() {
int x=0,f=1; char ch=getchar();
while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9') { x = (x<<3)+(x<<1)+(ch^48); ch=getchar(); }
return x * f;
}
缩短读入数据的时间，从前往后遍历，相同则p++，真想不明白这个思路一开始怎么没想到？？这样的题目以前做过不少啊。。。还是要真正理解吧
#include
using namespace std;
const int N = 1e5 + 7;
int n;
char ch[N][107],sss[10]="NowCoder";
inline int read() {
int x=0,f=1; char ch=getchar();
while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9') { x = (x<<3)+(x<<1)+(ch^48); ch=getchar(); }
return x * f;
}
int main()
{
n = read();
for(int i=1;i<=n;++i)
cin>>ch[i];
int p,Len;
for(int i=1;i<=n;++i) {
p = 0 ,Len = strlen(ch[i]);
for(int j=0;j 需要掌握：
字符类型二维数组。
eg：char a[n][m]表示n个长度不大于m的字符串，m的大小表示第n个字符串的第m个字符
scanf("%s",a[i]);//输入第i个字符串
char a[3][6]={“tread”,“micrp”,“soft”};
for(int i=0;i<3;i++)
{
printf("%s ",a[i]);//依次输出三个字符串
}
若固定i不变
a[i][j]代表的是第i个字符串的第j个字符
以下是一道题意晦涩其实很简单的题目。
One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 to n·k.
There were k children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the i-th (1 ≤ i ≤ k) child wrote the number ai (1 ≤ ai ≤ n·k). All numbers ai accidentally turned out to be different.
Now the children wonder, how to divide the orange so as to meet these conditions:
each child gets exactly n orange segments;
the i-th child gets the segment with number ai for sure;
no segment goes to two children simultaneously.
Help the children, divide the orange and fulfill the requirements, described above.
Input
The first line contains two integers n, k (1 ≤ n, k ≤ 30). The second line contains k space-separated integers a1, a2, …, ak (1 ≤ ai ≤ n·k), where ai is the number of the orange segment that the i-th child would like to get.
It is guaranteed that all numbers ai are distinct.
Output
Print exactly n·k distinct integers. The first n integers represent the indexes of the segments the first child will get, the second n integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.
You can print a child’s segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.
Examples
Input
2 2
4 1
Output
2 4
1 3
Input
3 1
2
Output
3 2 1
这个代码核心有些类似上一个题，虽然这两个题完全不沾边，但是同为水题，还是要做好朋友的
#include
#include
#include