Daily notes 2

1. In a N * N matrix, it is determined whether there is the diagonal element
used:

for(int i = 1; i <= n; i++){
    for(int j = i+1; j <= n; j++){
        if(abs(i-j) == abs(P[i]-P[j])){
            flag = false;
        }
    }
}

2. The input and output, on floating-point format
in scanf float is% f can, but must% lf double
in two possible are used printf% f

There is output format
printf ( "% m.nf");
represents a total of m bits, n bits after the decimal point, m is a decimal digit, sign bit
while a right-aligned +
- represented Left

2019/10/30 11:57

 //1.关于String中的find函数
 头文件：
 #include<cstring>
 #include<cstdio>
 #include<iostream>
 using namespace std;
 
 <1>返回查找字符的首字母下标
 string s1 = "qwer";
 string s2 = "er";
 int index = s1.find(s2);
 如果没有找到，就会返回nops
 if(index == string::nops){
     printf("没有找到该字符串！");
 }
 
 <2>返回子串出现在母串中的首次出现位置和最后一次出现的位置
 int index = s1.find_first_of(s2);
 int index = s1.find_last_of(s2);
 
 <3>查找某一给定位置的子串位置
 表示从字符串s1下标5开始查找，然后返回在s中的下标
 int index = s1.find(s2,5)
 
 <4>查找所有子串在母串中出现的位置
 //查找s 中flag 出现的所有位置。
    flag="a";
    position=0;
    int i=1;
    while((position=s.find(flag,position))!=string::npos)
    {
        cout<<"position  "<<i<<" : "<<position<<endl;
        position++;
        i++;
    }
    
 <5>反向查找子串在母串中出现的位置
 int index = s1.rfind(s2);