Tanya has nn candies numbered from 11 to nn . The ii -th candy has the weight aiai .
She plans to eat exactly n−1n−1 candies and give the remaining candy to her dad. Tanya eats candies in order of increasing their numbers, exactly one candy per day.
Your task is to find the number of such candies ii (let's call these candies good) that if dad gets the ii -th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days. Note that at first, she will give the candy, after it she will eat the remaining candies one by one.
For example, n=4n=4 and weights are [1,4,3,3][1,4,3,3] . Consider all possible cases to give a candy to dad:
- Tanya gives the 11 -st candy to dad (a1=1a1=1 ), the remaining candies are [4,3,3][4,3,3] . She will eat a2=4a2=4 in the first day, a3=3a3=3 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 4+3=74+3=7 and in even days she will eat 33 . Since 7≠37≠3 this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the 22 -nd candy to dad (a2=4a2=4 ), the remaining candies are [1,3,3][1,3,3] . She will eat a1=1a1=1 in the first day, a3=3a3=3 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 33 . Since 4≠34≠3 this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the 33 -rd candy to dad (a3=3a3=3 ), the remaining candies are [1,4,3][1,4,3] . She will eat a1=1a1=1 in the first day, a2=4a2=4 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 44 . Since 4=44=4 this case should be counted to the answer (this candy is good).
- Tanya gives the 44 -th candy to dad (a4=3a4=3 ), the remaining candies are [1,4,3][1,4,3] . She will eat a1=1a1=1 in the first day, a2=4a2=4 in the second day, a3=3a3=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 44 . Since 4=44=4 this case should be counted to the answer (this candy is good).
In total there 22 cases which should counted (these candies are good), so the answer is 22 .
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105 ) — the number of candies.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1041≤ai≤104 ), where aiai is the weight of the ii -th candy.
Print one integer — the number of such candies ii (good candies) that if dad gets the ii -th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days.
7 5 5 4 5 5 5 6
2
8 4 8 8 7 8 4 4 5
2
9 2 3 4 2 2 3 2 2 4
3
Note
In the first example indices of good candies are [1,2][1,2].
In the second example indices of good candies are [2,3][2,3].
In the third example indices of good candies are [4,5,9][4,5,9].
分析：题目大意，给你一个序列，删去一个数值之后，要求剩下序列奇数和偶数的和相同，问有多少种删法。
对数列做前缀和，奇加偶减，遍历每个位置，检测除去当前位置，前半部分前缀和和后半部分前缀和是否相等
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 5 const int maxn = 2*1e5+10; 6 int s[maxn]; 7 int ans=0; 8 9 int main(int argc, char const *argv[]) 10 { 11 int n; 12 cin>>n; 13 memset(s,0,sizeof(s)); 14 for( int i=1; i<=n; i++ ){ 15 int a; 16 cin>>a; 17 s[i]=s[i-1]+(i&1?a:-a); 18 } 19 20 for( int i=1; i<=n; i++ ){/*减去第i位置前半部分前缀和和后半部分前缀和是否相等*/ 21 if(s[i-1]==s[n]-s[i]) ans++; 22 } 23 cout<<ans<<endl; 24 return 0; 25 }