牛客多校暑假训练营2020 （第三场） B-【Classical String Problem】

B-Classical String Problem

题目描述：
Given a string S consists of lower case letters. You’re going to perform Q operations one by one. Each operation can be one of the following two types:

Modify: Given an integer x. You need to modify S according to the value of x. If x is positive, move the leftmost x letters in S to the right side of S; otherwise, move the rightmost |x| letters in S to the left side of S.
Answer: Given a positive integer x. Please answer what the x-th letter in the current string S is.

输出描述:

For each answer operation, please output a letter in a separate line representing the answer to the operation. The order of the output should match the order of the operations in the input.

输入：

nowcoder
6
A 1
M 4
A 6
M -3
M 1
A 1

输出：

n
o
w

备注：

Initially, S is ‘nowcoder’, six operations follow.
• The 1-st operation is asking what the 1-st letter is. The answer is ‘n’.
• The 2-nd operation is to move the leftmost 4 letters to the rightmost side, so S is modified to ‘odernowc’.
• The 3-rd operation is asking what the 6-th letter is. The answer is ‘o’.
• The 4-th operation is to move the rightmost 3 letters to the leftmost side, so S is modified to ‘owcodern’.
• The 5-th operation is to move the leftmost 1 letter to the rightmost side, so S is modified to ‘wcoderno’.
• The 6-th operation is asking what the 1-st letter is. The answer is ‘w’.

刚开始按照题目意思敲了个模拟，用string的加法和分割来做，但是会tttt，其实使用指针会更方便，我指针没太学好，就找一下规律，把字符串看成一个环，每次变形我们只需要改动这个字符串的初始位置，就可以了。
还有一个小问题：在输入字符串时，碰到了从键盘输入字符的问题。

对于scanf()而言，%c是个较为特殊的说明符。 %c前没空格，scanf()将读取标准输入流中的第一个字符，%c前有空格，scanf()则读取标准输入流中第一个非空白字符
为什么scanf的%c前为什么要加空格https://blog.csdn.net/u010850265/article/details/9136679

代码如下：

#include  //万能头文件
using namespace std;
const int maxn=20000006;

typedef long long ll;
char s[maxn];
int main(){
    scanf("%s",&s);
    int t,k=0,len;
    len=strlen(s);
    scanf("%d",&t);
    while(t--){
        char op;
        ll x;
        scanf(" %c %lld",&op,&x);//在%c前面加入空格跳过空白符
        if(op=='A') printf("%c\n",s[(k+len+x-1)%len]);
        else if(op=='M') k=(k+len+x)%len;
    }

     return 0;
}

**
附上大神的标准题解及源代码

#include
#define N 100005
using namespace std;
map<int,int>G;
int a[N],n;char s[N];
int main(){
	//freopen("test7.in","r",stdin);
	scanf("%d",&n);
	scanf("%s",s+1);
	for (int i=1;i<=n;i++)
		a[i]=a[i-1]+((s[i]=='1')?1:-1);
		
	//a[r]-a[l-1]==0
	int ans=0;
	G[0]=0;
	for (int i=1;i<=n;i++)
		if (G.find(a[i])!=G.end())
			ans=max(ans,i-G[a[i]]);
		else
			G[a[i]]=i;
	
	int zero=0,one=0;
	for (int i=1;i<=n;i++)
		if (s[i]=='0') zero++;else one++;
	printf("%d %d\n",ans,min(zero,one)*2);
}