Stas and the Queue at the Buffet

https://codeforces.com/contest/1151/problem/D

题意：

C++版本一

题解：

1、根据a和b的关系分类a>b的在前面，因为越前面，对a的权值越大，所以要使得a的值越小，同理使得越后面b越小；

2、如果1中大小关系相同，根据a，b的差值大小排序，代价问题，如果相邻两个交换位置，意味着前面一个+a-b,后面一个-a*b，根据1的思想分类以后，需要按差值大小排序；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{
    ll a,b;
    bool operator <(const node &S)const{
        if((a<b)==(S.a<S.b)){
            if(a<b){
                return b-a<S.b-S.a;
            }else{
                return a-b>S.a-S.b;
            }
        }
        return (a<b)<(S.a<S.b);
    }
}e[N];
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%I64d%I64d",&e[i].a,&e[i].b);
    }
    sort(e+1,e+n+1);
    for(int i=1;i<=n;i++){//cout<<e[i].a<<" "<<e[i].b<<endl;
        ans+=e[i].a*(i-1)+e[i].b*(n-i);
    }
    cout<<ans<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题解：对版本一的化简

因为一些简单的数学性质

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{
    ll a,b;
    bool operator <(const node &S)const{
        return b-a<S.b-S.a;
    }
}e[N];
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%I64d%I64d",&e[i].a,&e[i].b);
    }
    sort(e+1,e+n+1);
    for(int i=1;i<=n;i++){//cout<<e[i].a<<" "<<e[i].b<<endl;
        ans+=e[i].a*(i-1)+e[i].b*(n-i);
    }
    cout<<ans<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}