codeforces862E Mahmoud and Ehab and the function

把式子中的a和b拆開。
後面b的求和的部分是不變的。
前面對a的和的部分可能會+x，-x，不變。
每次在sumb裏面二分一個最接近的就好。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#define rep(i,j,k) for(int i = j; i <= k; i++)
#define dow(i,j,k) for(int i = j; i >= k; i--)
#define fi first
#define se second
using namespace std;


typedef long long ll;
typedef pair<int,int> pi;

const int N = 1e5 + 100;
int n, m, q;
ll suma[N], b[N], sumb[N];

ll solve(ll x) {
	int pos = lower_bound(sumb + n, sumb + m + 1, x) - sumb;
	if (pos > m) return x - sumb[m];
	if (pos == n) return sumb[pos] - x;
	return min(x - sumb[pos-1], sumb[pos] - x);
}

int main() {
	scanf("%d %d %d", &n, &m, &q);
	rep(i,1,n) {
		scanf("%lld", &suma[i]);
		if (i & 1) suma[i] = suma[i-1] + suma[i];
		else suma[i] = suma[i-1] - suma[i];
	}
	rep(i,1,m) {
		scanf("%lld", &b[i]);
	}
	rep(i,1,n) {
		if (i & 1) sumb[i] = sumb[i-1] + b[i];
		else sumb[i] = sumb[i-1] - b[i];
	}
	int f = -1;
	if (n & 1) f = 1;
	rep(i,1, m - n) {
		sumb[i + n] = -(sumb[i + n -1] - b[i]) + b[i + n] * f;
	}
	sort(sumb + n, sumb + m + 1);
	ll a = suma[n];
	printf("%lld\n", solve(a));
	rep(i,1,q) {
		int l, r, x;
		scanf("%d %d %d", &l, &r, &x);
		if ((r - l + 1) & 1) {
			if (l & 1) a += x;
			else a -= x;
		}
		printf("%lld\n", solve(a));
	}

}