codeforces E. Mahmoud and Ehab and the function（二分+思维）

题目链接：http://codeforces.com/contest/862/problem/E

题解：水题显然利用前缀和考虑一下然后就是二分b的和与-ans_a最近的数（ans_a表示a的前缀和（奇加偶减））

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const int M = 1e5 + ;
ll a[M] , b[M] , sum[M] , ans_b[M];
int main() {
    int n , m , q;
    scanf("%d%d%d" , &n , &m , &q);
    for(int i =  ; i <= n ; i++) scanf("%lld" , &a[i]);
    for(int i =  ; i <= m ; i++) scanf("%lld" , &b[i]);
    sum[] = ;
    for(int i =  ; i <= m ; i++) {
        if(i % ) {
            sum[i] = sum[i - ] + b[i];
        }
        else {
            sum[i] = sum[i - ] - b[i];
        }
    }
    for(int i =  ; i <= m - n +  ; i++) {
        if(!(i % )) {
            ans_b[i - ] = sum[i + n - ] - sum[i - ];
        }
        else {
            ans_b[i - ] = sum[i - ] - sum[i + n - ];
        }
    }
    sort(ans_b , ans_b + m - n + );
    ll ans_a = ;
    for(int i =  ; i <= n ; i++) {
        if(i % ) {
            ans_a += a[i];
        }
        else {
            ans_a -= a[i];
        }
    }
    int pos = ;
    pos = lower_bound(ans_b , ans_b + m - n +  , -ans_a) - ans_b;
    pos = min(m - n , pos);
    printf("%lld\n" , min(abs(ans_a + ans_b[max( , pos - )]) , min(abs(ans_a + ans_b[pos]) , abs(ans_a + ans_b[min(m - n , pos + )]))));
    while(q--) {
        int l , r;
        ll x;
        scanf("%d%d%lld" , &l , &r , &x);
        if(l % ) {
            if((r - l + ) % ) {
                ans_a += x;
            }
        }
        else {
            if((r - l + ) % ) {
                ans_a -= x;
            }
        }
        pos = lower_bound(ans_b , ans_b + m - n +  , -ans_a) - ans_b;
        pos = min(m - n , pos);
        printf("%lld\n" , min(abs(ans_a + ans_b[max( , pos - )]) , min(abs(ans_a + ans_b[pos]) , abs(ans_a + ans_b[min(m - n , pos + )]))));
    }
    return ;
}