A - Sherlock Bones Gym - 101350A

题意:The great dog detective Sherlock Bones is on the verge of a new discovery. But for this problem, he needs the help of his most trusted advisor -you- to help him fetch the answer to this case.

He is given a string of zeros and ones and length N.

Let F(x, y) equal to the number of ones in the string between indices x and yinclusively.

Your task is to help Sherlock Bones find the number of ways to choose indices (i, j, k) such that i < j < ksj is equal to 1, and F(i, j) is equal to F(j, k).

思路:这个题其实就可以转化为求区间的个数,其中区间内1的个数为奇数,其实思路还是挺好转化的,但是个人感觉代码写起来技巧性还是很强的。

先根据输入的串s求出区间内1个数的奇偶(根据异或),然后在倒着统计有奇数个1的区间个数,偶数个1的区间个数,最后统计总区间个数,在统计总区间个数的时候注意统计方法,同时注意对字符串的处理方式还有计数方式。

#include 
#include
#include
#include
#include
#include
#include
#include
#include 
#include 

using namespace std;
const int maxn=2e5+20;
int per[maxn];
int g[maxn][2];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        memset(per,0,sizeof(per));
        memset(g,0,sizeof(g));
        int n;
        cin>>n;
        string s;
        cin>>s;
        for(int i=0; i=0; i--)
        {
            if(per[i]==1)
            {
                g[i][1]=g[i+1][1]+1;
                g[i][0]=g[i+1][0];
            }
            else
            {
                g[i][0]=g[i+1][0]+1;
                g[i][1]=g[i+1][1];
            }
        }
        s="0"+s;
        int flag=0;
        long long ans=0;
        int last=n+1;
        for(int i=n; i>0; i--)
        {
            if(!flag)
            {
                if(s[i]=='1')
                {
                    flag=1;
                    last=i;
                }
            }
            else
            {
                if(per[i-1]==0)
                {
                    ans+=g[last+1][1];
                }
                else
                {
                    ans+=g[last+1][0];
                }
                if(s[i]=='1')
                {
                    last=i;
                }
            }
        }
        cout<


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