ICPC Mid-Central Regional 2019 L. Sum and Product

2020.5.3
题目链接：传送门
题目概述：查找1-n个不为零的数列中和和差相等的子序列个数
解题思路：这道题用暴力非常简单，但肯定超时，一直在找优化的办法。首先想到和和积相等，肯定是在考察数组中1和2的数量，之前用1和2分治卡过去了，今天上kattis看貌似已经不行。得另想办法，但是我想不到啊，分治好像也不行，没有优化空间。然后问了下群里西工大的大佬@I，给出关键思路，即当暴力的时候，如果 ai 大于1，则超过log2个不为1的数字之后和永远无法和积相等。妙啊，然后因为坑爹的2e6数据re了一波。然后AC。看来以后还是得多练一下思路。区域赛目前补出来9道了，还是得多加练习，争取今年去一趟nac。
特别鸣谢：凯哥
代码：

#include 
using namespace std;
#define limit (100000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%lld\n",a );
#define pi(a,b) pair
#define rep(i, a, b) for(int i = a ; i <= b ; ++i)
#define per(i, a, b) for(int i = b ; i >= a ; --i)
#define mint(a,b,c) min(min(a,b), c)
#define MOD 998244353
#define FOPEN freopen("C:\\Users\\administrator01\\CLionProjects\\untitled24\\data.txt", "rt", stdin)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n , m;
int vis[limit], f[limit];
void check(int x,  int &ans){
    int t = 1;
    int len = n;
    ll prod = f[x], sum = f[x];
    int lim = log2(n - x + 1) + 1;
    int s = 0;
    for(int i = x + 1;  t <= lim && s <= lim && i <= n ; ){
            if(!vis[i]){
                ++t;
                prod *= f[i];
                sum += f[i];
                if(prod == sum){
                    ++ans;
                    //printf("At %d, %dth, prod %lld and sum %lld\n", x,i,prod, sum);
                }
                ++i;
            }else{
                ++s;
                ++sum;
                if(prod >= sum && prod - sum < vis[i]){
                    //printf("At %d, %dth\n", x,i);
                    ++ans;
                }
                sum += vis[i] - 1;
                i += vis[i];
            }

    }
}

int main(){
#ifdef LOCAL
    FOPEN;
    //freopen("C:\\Users\\administrator01\\CLionProjects\\untitled24\\out.txt", "w", stdout);
#endif
    n = read();
    int flag = 0, iter = 0, num = 0;
    int s1 = 0, s2 = 0;
    rep(i ,1, n){
        f[i] = read();
        if(f[i] == 1){
            if(flag){
                vis[iter]++;
                continue;
            }
            flag = 1;
            iter = i;
            vis[iter]++;
            ++num;
            s1 = 1;
        }
        if(f[i] == 2){
            s2++;
        }
        if(flag && f[i] != 1){
            num = iter = flag = 0;//归零
        }
    }
    rep(i ,1, n){
        if(vis[i - 1])vis[i] = vis[i - 1] - 1;
    }
    int ans = 0;
    for(int i = 1 ; i <= n ; ++i){
        check(i,ans);
    }
    write(ans);
    return 0;
}