AtCoder Beginner Contest 081 D

D Non-decreasing

题意：给N个数，有操作f(a,b) 把第a个数的值加到第b个数上。求一个小于2N的操作序列，使给出的序列变成非递减序列。

题解:当序列为非负序列时，通过a[i+1]+=a[i]构造；

当为非正序列时a[i]+=a[i+1];

所以先用N个操作(加绝对值最大的数)变成非正/非负序列，然后按照上面的构造方法构造即可。

注意边界。

#include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef long double db;
    typedef pair<int,int> pii;
    typedef vector<int> vi;
    #define de(x) cout << #x << "=" << x << endl
    #define rep(i,a,b) for(int i=a;i<(b);++i)
    #define per(i,a,b) for(int i=b-1;i>=(a);i--) 
    #define all(x) (x).begin(),(x).end()
    #define sz(x) (int)(x).size()
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define lb(x) (x&-(x))
    const int N = ;
    int a[];
    int main(){
        int maxx=INT_MIN;
        int maxxi,minni;
        int minn=INT_MAX;
        int n;
        cin>>n;
        rep(i,,n) 
        {
            scanf("%d",&a[i]);
            if(a[i]>maxx)
            {
                maxx=a[i];
                maxxi=i;
            }
            if(a[i]<minn)
            {
                minn=min(minn,a[i]);
                minni=i;
            }
        }


        printf("%d\n",n+n-);
        if(maxx+minn>=)
        {    
             rep(i,,n)
             {
                printf("%d %d\n",maxxi+,i+);
                a[i]+=maxx;
             }

    //       rep(i,0,n) cout<<a[i]<<" ";cout<<endl;
            rep(i,,n-)
            {
                printf("%d %d\n",i+,i+);
                a[i+]+=a[i];
            }
        }
        else
        {
             rep(i,,n)
             {
                printf("%d %d\n",minni+,i+);
                a[i]+=minn;
             }      
    //       rep(i,0,n) cout<<a[i]<<" ";cout<<endl;     
            per(i,,n-)
            {
                printf("%d %d\n",i+,i+);
                a[i]+=a[i+];
            }
        }



        return ;
    }