标签：多校

Average

There are n soda sitting around a round table. soda are numbered from 1 to n and i-th soda is adjacent to (i+1)-th soda, 1-st soda is adjacent to n-th soda.

Each soda has some candies in their hand. And they want to make the number of candies the same by doing some taking and giving operations. More specifically, every two adjacent sodax and y can do one of the following operations only once:
1. x-th soda gives y-th soda a candy if he has one;
2. y-th soda gives x-th soda a candy if he has one;
3. they just do nothing.

Now you are to determine whether it is possible and give a sequence of operations.

There are multiple test cases. The first line of input contains an integerT, indicating the number of test cases. For each test case:

The first contains an integer n(1≤n≤105), the number of soda.
The next line contains n integers a1,a2,…,an(0≤ai≤109), where ai denotes the candy i-th soda has.

For each test case, output "YES" (without the quotes) if possible, otherwise output "NO" (without the quotes) in the first line. If possible, then the output an integerm(0≤m≤n) in the second line denoting the number of operations needed. Then each of the followingm lines contain two integers x and y(1≤x,y≤n), which means that x-th soda gives y-th soda a candy.

3
6
1 0 1 0 0 0
5
1 1 1 1 1
3
1 2 3

NO
YES
0
YES
2
2 1
3 2

#include<stdio.h>
#include<string.h>
#define ll __int64
const int N  = 100005;
struct EDG{
    int x,y;
};
int n,step;
ll a[N],b[N],avg;
EDG edg[N<<1];
void addEdg(int x,int y)
{
    edg[step].x=x;
    edg[step].y=y;
    step++;
}
bool findout(int l)
{
    while(l<=n){
        if(a[l]==avg){l++;continue;}
        if(l==n)return 0;
        if(a[l]>avg){
            if(a[l]-avg>1){
                return 0;
            }
            else{
                l++;
                a[l]++;
                a[l-1]--;
                addEdg(l-1 , l); //从左向右过一个
            }
        }
        else {
            if(avg-a[l]>1){
                return 0;
            }
            if(a[l+1]<avg){
                return 0;
            }
            l++; a[l]--; a[l-1]++;  
            addEdg(l , l-1); //从右向左过一个
        }
    }

    return 1;
}
void print()
{
    printf("YES\n%d\n",step);
    for(int i=0; i<step; i++)
        printf("%d %d\n",edg[i].x,edg[i].y);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        avg=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%I64d",&b[i]);
            avg+=b[i] ;
        }
        if(avg%n!=0){
            printf("NO\n");continue;
        }
        avg/=n;
        step=0;
//因是一个圈，所以从第1 个分情况讨论，注意每次求完一轮把step 初始为0，我就是在这个WA了好多次
        if(b[1]-avg==2){
            for(int i=1; i<=n; i++)
            a[i]=b[i];
            a[1]--; a[2]++; addEdg(1 , 2);
            if(n>2){
                a[1]--;a[n]++; addEdg(1, n);
                if( findout(1) ){
                    print(); continue;
                }
            }
        }
        else if(b[1]-avg==-2){
            if(b[2]&&b[n]){
                for(int i=1; i<=n; i++)
                a[i]=b[i];
                a[1]++; a[2]--; addEdg(2 , 1);
                if(n>2){
                    a[1]++;a[n]--; addEdg(n, 1);
                    if( findout(1) ){
                        print(); continue;
                    }
                }
            }
        }
        else if(b[1]-avg==1){
            for(int i=1; i<=n; i++)
            a[i]=b[i];
            a[1]--; a[2]++; addEdg(1 , 2);
            if( findout(1) ){
                print(); continue;
            }
            step=0;
            for(int i=1; i<=n; i++)
            a[i]=b[i];
            a[1]--; a[n]++; addEdg(1 , n);
            if( findout(1) ){
                print(); continue;
            }
        }
        else if(b[1]-avg==-1){
            if(b[2]){
                for(int i=1; i<=n; i++)
                a[i]=b[i];
                a[1]++; a[2]--; addEdg(2 , 1);
                if( findout(1) ){
                    print(); continue;
                }
            }
            step=0;
            if(b[n]){
                for(int i=1; i<=n; i++)
                a[i]=b[i];
                a[1]++; a[n]--; addEdg(n , 1);
                if( findout(1) ){
                    print(); continue;
                }
            }
        }
        else if(avg==b[1]){
            for(int i=1; i<=n; i++)
            a[i]=b[i];
            if( findout(1) ){
                print(); continue;
            }
            step=0;
            if(b[2]){
                for(int i=1; i<=n; i++)
                a[i]=b[i];
                a[n]++; a[2]--; addEdg(1 , n); addEdg(2 , 1);
                if( findout(1) ){
                    print(); continue;
                }
            }
            step=0;
            if(b[n]){
                for(int i=1; i<=n; i++)
                a[i]=b[i];
                a[n]--; a[2]++;addEdg(1 , 2); addEdg(n , 1);
                if( findout(1) ){
                    print(); continue;
                }
            }
        }
        printf("NO\n");
    }
}

HDU 5353 Average（平分值，求步聚）多校6

标签：多校

原文地址：http://blog.csdn.net/u010372095/article/details/47394977