Binary Differences

https://csacademy.com/contest/archive/task/binary-differences

n个数,只有0和1,求所有子区间价值不相同的有多少中,价值是0的个数-1的个数

解法:0的贡献是1,1的贡献是-1,求出贡献的前缀和为s[i],利用上一个区间[l,r]求出当前区间[sum-r,sum-l],同时更新最大范围

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pii pair<int,int>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-12;
const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

int a[N];
int main()
{
    int n,sum=0;
    scanf("%d",&n);
    int l,r,tel,ter;
    l=r=tel=ter=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(a[i])sum--;
        else sum++;
        l=min(l,sum-ter);
        r=max(r,sum-tel);
        tel=min(tel,sum);
        ter=max(ter,sum);
    }
    printf("%d
",r-l+1);
    return 0;
}
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