Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Sample Input 1
8 3
ACACTACG
3 7
2 3
1 8
Sample Output 1
2
0
3
Query 1: the substring of S starting at index 3 and ending at index 7 is ACTAC. In this string, AC occurs twice as a substring.
Query 2: the substring of S starting at index 2 and ending at index 3 is CA. In this string, AC occurs zero times as a substring.
Query 3: the substring of S starting at index 1 and ending at index 8 is ACACTACG. In this string, AC occurs three times as a substring.
思路:
将一个做这道题的主思路,举个例子吧,例子容易理解:a[1]=1;a[2]=1+2:a[3]=1+2+3……a[n]=1+2+3+……+n;那么求第四个到底二十个的和怎么求?先明白一个简单的例子:从2到6之间的和是a[6]-a[2]=1+2+3+4+5+6-(1+2)+2=3+4+5+6+2,这个例子看懂的话,那么上一个例子是不是就明白了,a[20]-a[4]+4就是从20到4之间的和。那么意明白,接下来这道题就好做了。
代码:
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int main() {
long long n,m;string s;
while(cin>>n>>m>>s){
int a[200000],v=0;memset(a,0,sizeof(a));
for(int i=0;i<s.length();i++){
a[i]=v;
if(s[i]=='A'&&s[i+1]=='C'){
a[i+1]=v+1;i++;v++;
}
}
cout<<endl;
for(int i=0;i<m;i++){
int x,y;cin>>x>>y;
cout<<a[y-1]-a[x-1]<<endl;
}
}
return 0;
}