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 + 10;
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 = 1 ; i <= n ; i++) scanf("%lld" , &a[i]);
    for(int i = 1 ; i <= m ; i++) scanf("%lld" , &b[i]);
    sum[0] = 0;
    for(int i = 1 ; i <= m ; i++) {
        if(i % 2) {
            sum[i] = sum[i - 1] + b[i];
        }
        else {
            sum[i] = sum[i - 1] - b[i];
        }
    }
    for(int i = 1 ; i <= m - n + 1 ; i++) {
        if(!(i % 2)) {
            ans_b[i - 1] = sum[i + n - 1] - sum[i - 1];
        }
        else {
            ans_b[i - 1] = sum[i - 1] - sum[i + n - 1];
        }
    }
    sort(ans_b , ans_b + m - n + 1);
    ll ans_a = 0;
    for(int i = 1 ; i <= n ; i++) {
        if(i % 2) {
            ans_a += a[i];
        }
        else {
            ans_a -= a[i];
        }
    }
    int pos = 0;
    pos = lower_bound(ans_b , ans_b + m - n + 1 , -ans_a) - ans_b;
    pos = min(m - n , pos);
    printf("%lld\n" , min(abs(ans_a + ans_b[max(0 , pos - 1)]) , min(abs(ans_a + ans_b[pos]) , abs(ans_a + ans_b[min(m - n , pos + 1)]))));
    while(q--) {
        int l , r;
        ll x;
        scanf("%d%d%lld" , &l , &r , &x);
        if(l % 2) {
            if((r - l + 1) % 2) {
                ans_a += x;
            }
        }
        else {
            if((r - l + 1) % 2) {
                ans_a -= x;
            }
        }
        pos = lower_bound(ans_b , ans_b + m - n + 1 , -ans_a) - ans_b;
        pos = min(m - n , pos);
        printf("%lld\n" , min(abs(ans_a + ans_b[max(0 , pos - 1)]) , min(abs(ans_a + ans_b[pos]) , abs(ans_a + ans_b[min(m - n , pos + 1)]))));
    }
    return 0;
}