Codeforces Round #376 (Div. 2) F 区间求和+二分

本文介绍了一道CodeForces竞赛平台上的F题的解决思路。题目要求在给定数组中选择一个数作为基准，计算所有数转换为该数倍数后的最大总和。文章通过排序和二分查找的方法实现了高效求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目传送门： http://codeforces.com/contest/731/problem/F

题意：给n个数(n<=20W)，从中找一个数，然后在数组中将值为该数倍数的数全部加起来，如果不是该数的倍数，则减去一定的数使其变为该数的倍数。然后找最大的和

思路：先排个序，从第一个数开始遍历，从k=1开始找在[a*k,a*k+1)之间的数目j,a*k*j即为该段数的和，查找时可以用二分

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   ll            long long
#define   ull           unsigned long long
#define   mem(n,v)      memset(n,v,sizeof(n))
#define   MAX           200005
#define   MAXN          1000005
#define   PI            3.1415926
#define   E             2.718281828459
#define   opnin         freopen("text.in.txt","r",stdin)
#define   opnout        freopen("text.out.txt","w",stdout)
#define   clsin         fclose(stdin)
#define   clsout        fclose(stdout)
#define   haha1         cout << "haha1"<< endl
#define   haha2         cout << "haha2"<< endl
#define   haha3         cout << "haha3"<< endl

const int    INF    =   0x3f3f3f3f;
const ll     INFF   =   0x3f3f3f3f3f3f3f3f;
const double pi     =   3.141592653589793;
const double inf    =   1e18;
const double eps    =   1e-8;
const ll     mod    =   1e9+7;
const ull    mx     =   133333331;

ll a[MAX];
int vis[MAX];
int main()
{
    mem(a,0);
    mem(vis,0);
    int n;
    cin >> n;
    for(int i=1;i<=n;i++){
        scanf("%I64d",&a[i]);
    }
    sort(a+1,a+n+1);
    ll sum = 0,sum1;
    for(int i=1;i<=n;i++){
        if(vis[a[i]]) continue;
        vis[a[i]] = 1;
        sum1 = 0;
        int x = 0,y = 1;
        for(ll j=a[i];j<=a[n];j+=a[i]){
            x = lower_bound(a+y,a+n+1,j) - a;
            y = lower_bound(a+y,a+n+1,j+a[i]) - a;
//            cout << j << ' ' << x << ' ' << y << endl;
            sum1 += j * (y-x);
//            cout << sum1 << endl;
        }
//        cout << endl;
        sum = max(sum,sum1);

    }
    cout << sum << endl;
}