AtCoder Regular Contest 119 C - ARC Wrecker 2（同余定理+思维）

Problem Statement

There are $N$ buildings along the AtCoder Street, numbered $1$ through $N$ from west to east. Initially, Buildings $1,2,\dots ,N$ have the heights of $A_1,A_2,\dots ,A_N$, respectively.

Takahashi, the president of ARC Wrecker, Inc., plans to choose integers $l$ and $r$ $(1\le l<r\le N)$ and make the heights of Buildings $l,l+1,\dots ,r$ all zero. To do so, he can use the following two kinds of operations any number of times in any order:

• Set an integer $x$ $(l\le x\le r-1)$ and increase the heights of Buildings $x$ and $x+1$ by $1$ each.
• Set an integer $x$ $(l\le x\le r-1)$ and decrease the heights of Buildings $x$ and $x+1$ by $1$ each. This operation can only be done when both of those buildings have heights of $1$ or greater.

Note that the range of $x$ depends on $(l,r)$.

How many choices of $(l,r)$ are there where Takahashi can realize his plan?

Constraints

• $2\le N\le 300000$
• $1\le A_i\le {10}^9$ $(1\le i\le N)$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$
$A_1$ $A_2$ $\cdots$ $A_N$

Output

Print the answer.

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Sample Input 1

5
5 8 8 6 6

Sample Output 1

3

Takahashi can realize his plan for $(l,r)=(2,3),(4,5),(2,5)$.

For example, for $(l,r)=(2,5)$, the following sequence of operations make the heights of Buildings $2,3,4,5$ all zero.

• Decrease the heights of Buildings $4$ and $5$ by $1$ each, six times in a row.
• Decrease the heights of Buildings $2$ and $3$ by $1$ each, eight times in a row.

For the remaining seven choices of $(l,r)$, there is no sequence of operations that can realize his plan.

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Sample Input 2

7
12 8 11 3 3 13 2

Sample Output 2

3

Takahashi can realize his plan for $(l,r)=(2,4),(3,7),(4,5)$.

For example, for $(l,r)=(3,7)$, the following figure shows one possible solution.

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Sample Input 3

10
8 6 3 9 5 4 7 2 1 10

Sample Output 3

1

Takahashi can realize his plan for $(l,r)=(3,8)$ only.

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Sample Input 4

14
630551244 683685976 249199599 863395255 667330388 617766025 564631293 614195656 944865979 277535591 390222868 527065404 136842536 971731491

Sample Output 4

8

题意：
有 n 个数，每次可以将 a[x],a[x+1] 同时 +1/-1 ，问存在多少区间同时减为 0

题解：
直觉告诉我用差分来做，然后模拟出来的结果时间复杂度都是 O(n^2)
直到看到了 hu_tao 大佬的讨论，一语惊醒，每次操作一定是将一个奇数位和偶数位同时操作，所以无论怎么变一个区间想要同时变为 0，那么这个区间上奇数位置和偶数位置的加和是相同的；
所以将奇数位置处的值置为负数，利用同余定理，就可以找到任意一个连续的子段加和为 0

const int N=1e6+5;
 
    int n, m, _;
    int i, j, k;
    ll a[N];
    map<ll,int> mp;

signed main()
{
    //IOS;
    while(~sd(n)){
        for(int i=1;i<=n;i++){
            sll(a[i]);
            if(i%2==0) a[i]=-a[i];
        }
        ll ans=0;
        mp[0]++;
        for(int i=1;i<=n;i++){
            a[i]+=a[i-1];
            if(mp[a[i]]) ans+=mp[a[i]];
            mp[a[i]]++;
        }
        pll(ans);
        mp.clear();
    }
    //PAUSE;
    return 0;
}