# 3724. 求和公式

https://bbs.csdn.net/topics/392963885 老哥，看看这个会不会有思路

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,q;
ll a,b,c,d;
ll ans;
int main()
{
cin>>n>>q;
while(q--){
cin>>a>>b>>c>>d;
ans=max(0LL,min(c,d) - max(a,b) + 1) + max(0LL,min(c,n - 1 - b) - max(a,n - 1 - d) + 1);
if(n % 2 && n / 2 >= a && n / 2 <= c && n / 2 >= b && n / 2 <= d)--ans;
cout<<ans<<endl;
}
return 0;
}

#include <iostream>
using namespace std;

typedef long long LL;
LL n, q, x1, y1, x2, y2, sum;

int main() {
cin >> n >> q;
for (int i = 1; i <= q ; i++) {
sum = 0;
cin >> x1 >> y1 >> x2 >> y2;
x1++, y1++, x2++, y2++;

// 统计对角线
if (x1 <= y1) {
if (x1 != y1) {
if (x2 - y1 >= 0) sum += min(x2, y2) - y1 + 1;
}
if (x1 == y1) {
sum += min(x2, y2) - y1 + 1;
}
} else {
if (min(x2, y2) - x1 >= 0) sum += min(x2, y2) - max(x1, y1) + 1;
}

// 统计反对角线
LL tmp = y1;
y1 = n + 1 - y2;
y2 = n + 1 - tmp;

if (x1 <= y1) {
if (x1 != y1) {
if (x2 - y1 >= 0) sum += min(x2, y2) - y1 + 1;
}
if (x1 == y1) {
sum += min(x2, y2) - y1 + 1;
}
} else {
if (min(x2, y2) - x1 >= 0) sum += min(x2, y2) - max(x1, y1) + 1;
}

// 处理中心点是否被重复计算
LL mid = n / 2 + 1;
if (n & 1 == 1 && (mid >= x1 && mid <= x2 && mid >= n + 1 - y2 && mid <= n + 1 - y1)) sum--;

cout << sum << endl;
}

return 0;
}


#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
ios::sync_with_stdio(false);
ll n, q;
cin >> n >> q;
while (q--) {
ll x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
ll m1, s1, mainLine, subLine;
//主对角线
m1 = x1 - y1;
if (m1 < 0) {
mainLine = min(x2 - x1 + 1 + m1, y2 - y1 + 1);
}
else if (m1 > 0) mainLine = min(y2 - y1 + 1 - m1, x2 - x1 + 1);
else mainLine = min(x2 - x1 + 1, y2 - y1 + 1);
if (mainLine < 0) mainLine = 0;
//副对角线
s1 = n - 1 - x1 - y2;
if (s1 < 0) {
subLine = min(y2 - y1 + 1 + s1, x2 - x1 + 1);
}
else if (s1 > 0) subLine = min(x2 - x1 + 1 - s1, y2 - y1 + 1);
else subLine = min(x2 - x1 + 1, y2 - y1 + 1);
if (subLine < 0) subLine = 0;
ll ans = subLine + mainLine;
//中心是否重复计算
if ((n - 1) % 2 == 0 && x1 <= (n - 1) / 2 && x2 >= (n - 1) / 2 && y1 <= (n - 1) / 2 && y2 >= (n - 1) / 2) {
if (ans == 0) ans = 0;
else ans--;
printf("%lld\n", ans);
}
else printf("%lld\n", ans);
}
return 0;
}