2015多校6

Solved	Pro.ID	Title	Author	Source	(AC/Submit)Ratio
	5353	Average		2015 Multi-University Training Contest 6	(433/1704)25.41%
	5354	Bipartite Graph		2015 Multi-University Training Contest 6	(29/137)21.17%
	5355	Cake		2015 Multi-University Training Contest 6	(301/1709)17.61%
	5356	Deal		2015 Multi-University Training Contest 6	(16/43)37.21%
	5357	Easy Sequence		2015 Multi-University Training Contest 6	(75/285)26.32%
	5358	First One		2015 Multi-University Training Contest 6	(428/1406)30.44%
	5359	Group		2015 Multi-University Training Contest 6	(1/1)100.00%
	5360	Hiking		2015 Multi-University Training Contest 6	(443/848)52.24%
	5361	In Touch		2015 Multi-University Training Contest 6	(247/919)26.88%
	5362	Just A String		2015 Multi-University Training Contest 6	(93/385)24.16%
	5363	Key Set		2015 Multi-University Training Contest 6	(401/651)61.60%

 

HDU5353

n个人成环，每个人有a[i]块糖

相邻两个人x,y之间有且只有其中一种操作：

1 x给y一颗糖

2 y给x一颗糖

3 do nothing

求能否使得每个人手中糖的数量一样，如果可以任意一种输出步骤，spj

只要前两人的操作定下来后，后面的人的操作就定下来了，因为每条边只能有一种状态。3个for循环搞定。O(n)

#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
#define maxn 100010
int n;
int s[maxn],a[maxn];
int avg;
int cnt;
bool run()
{
    int i=2;
    int flag=1;
    while(i!=2||flag)
    {
        flag=0;

        int pre=(i-1+n)%n;
        if(a[pre]==avg);
        else if(a[pre]==avg+1)
        {
            a[pre]--;
            a[i]++;
            cnt++;
        }
        else if(a[pre]==avg-1)
        {
            a[pre]++;
            a[i]--;
            cnt++;
        }
        else return false;

        i=(i+1)%n;
    }
    return true;
}
void print()
{
    int i=2;
    int flag=1;
    while(i!=2||flag)
    {
        flag=0;
        int pre=(i-1+n)%n;
        if(s[pre]==avg);
        else if(s[pre]==avg+1)
        {
            s[pre]--;
            s[i]++;
            printf("%d %d\n",pre+1,i+1);
        }
        else if(s[pre]==avg-1)
        {
            s[pre]++;
            s[i]--;
            printf("%d %d\n",i+1,pre+1);
        }
        i=(i+1)%n;
    }
    return ;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ll sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&s[i]);
            sum+=s[i];
        }
        avg=sum/n;
        if(n==1)
        {
            printf("YES\n0\n");
            continue;
        }
        if(sum%n!=0)
        {
            printf("NO\n");
            continue;
        }
        //0--1
        cnt=0;
        for(int i=0;i<n;i++)a[i]=s[i];
        if(run())
        {
            printf("YES\n%d\n",cnt);
            print();
            continue;
        }

        //0->1
        cnt=1;
        for(int i=0;i<n;i++)a[i]=s[i];
        a[0]++;
        a[1]--;
        if(run())
        {
            s[0]++;
            s[1]--;
            printf("YES\n%d\n2 1\n",cnt);
            print();
            continue;
        }
        //0<-1;
        cnt=1;
        for(int i=0;i<n;i++)a[i]=s[i];
        a[0]--;
        a[1]++;
        if(run())
        {
            s[0]--;
            s[1]++;
            printf("YES\n%d\n1 2\n",cnt);
            print();
            continue;
        }
        puts("NO");
    }
    return 0;
}