[leetcode-38]Count and Say

自动生成序列的一道题，想起来昨天比赛里遇见的另一道题，问C_∞序列第N位是多少，昨天死磕到最后也就是写了个垃圾代码，然后输入N>10就不能运行，觉得这类题的要点是要开两个序列，一个生成另一个后，然后拷贝回来，然后记得往后推测的两个条件，

1.注意现有数列边界，不要越界，例如下面代码中的i+1<res.size()

2.注意题目限制推到第X位，例如Count and Say里要推到第N个序列，就有一个while(n--)的限制，昨天的垃圾代码里就有一个i<n的限制

然后上代码，先上Count and Say标程（题目没看懂直接看了标程...）：

string countAndSay(int n) {
if (n == 0) return "";
string res = "1";
while (--n) {
string cur = "";
for (int i = 0; i < res.size(); i++) {
int count = 1;
while ((i + 1 < res.size()) && (res[i] == res[i + 1])){
count++;
i++;
}
cur += to_string(count) + res[i];
}
res = cur;
}
return res;
}然后是昨天的题目[leetcode-481]Magic String：
A magical string S consists of only '1' and '2' and obeys the following rules:

The string S is magical because concatenating the number of contiguous occurrences of characters '1' and '2' generates the string S itself.

The first few elements of string S is the following: S = "1221121221221121122……"

If we group the consecutive '1's and '2's in S, it will be:

1 22 11 2 1 22 1 22 11 2 11 22 ......

and the occurrences of '1's or '2's in each group are:

1 2 2 1 1 2 1 2 2 1 2 2 ......

You can see that the occurrence sequence above is the S itself.

Given an integer N as input, return the number of '1's in the first N number in the magical string S.

Note: N will not exceed 100,000.附垃圾代码供参考，题解稍后我改改再发：
int magicalString(int n) {
int s[3125],k=2,countx=1,j,i,flag=0;
//k代表递推出来的位置
//countx代表第k位前1的个数
//i，j代表正在从（32*j+i）递推
s[0]=1;
for(i=2,j=0;k<n;i++){
if(i==32){
i=0;j++;
}
if(s[j]&1<<i==1){ //当前连续序列长度为1时
if(flag==1) //当前序列当1时
{
s[k/32]+=1<<k%32;
countx+=1;
}
}
else{ //当前连续序列长度为2时
if(flag==1) //当前序列应当添1时
{
s[k/32]+=1<<k%32;
k++;
s[k/32]+=1<<k%32;
countx+=2;
}
else{
k+=1;
}
}
k+=1;
flag=1-flag;
}
cout<<s[0]<<endl;
return countx;
}