CF18 C. Stripe

题目传送门：https://codeforces.com/problemset/problem/18/C

题目大意：
给你一串长度为(A)的序列，问有多少种分割方法，使得两段数字和相同

---

记前缀和(Pre)，记后缀和(Suf)，求所有(Pre_i=Suf_{i+1})的点即可

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
int Pre[N+10],Suf[N+10],A[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read(0),Ans=0;
	for (int i=1;i<=n;i++)	A[i]=read(0);
	for (int i=1;i<=n;i++)	Pre[i]=Pre[i-1]+A[i];
	for (int i=n;i>=1;i--)	Suf[i]=Suf[i+1]+A[i];
	for (int i=1;i<n;i++)	Ans+=(Pre[i]==Suf[i+1]);
	printf("%d
",Ans);
	return 0;
}