2015 多校联赛 ——HDU5353（构造）

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 soda x 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.

Sample Input

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

Sample Output

NO
YES
0
YES
2
2 1
3 2

对相邻两个数之间进行以下三种操作的一种,最后使他们相等

①a++ b-- ②a-- b++ ③nothing

如果(sum%n != 0)，直接失败。用一个数组来记录数与平均数之间的差值。

先枚举第一位数的三种情况，

当C[i] == 1时，从C[i+1]取一；

当C[i] == -1时，给C[i+1]一个；

当C[i] == 0时，nothing；

else：false。

ps：完全没想到要对第一位进行枚举 OoO，一直wa

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
const int mod = 1000000007;
int a[100050];
int aver,flag,n;
int p[100050][2];
int c[100050];
int tot;

bool solve()
{
for(int i = 3; i <= n; i++)
c[i] = a[i];
for(int i = 2; i <= n; i++)
{
if(c[i] == 0)
continue;
else if(c[i] == 1 && i != n)
{
c[i+1]++;
c[i]--;
p[tot][0] = i;
p[tot++][1] = i+1;
}
else if(c[i] == 1 && i == n)
{
c[1]++;
c[i]--;
p[tot][0] = i;
p[tot++][1] = 1;
}
else if(c[i] == -1 && i!= n)
{
c[i]++;
c[i+1]--;
p[tot][0] = i+1;
p[tot++][1] = i;
}
else if(c[i] == -1 && i== n)
{
c[i]++;
c[1]--;
p[tot][0] = 1;
p[tot++][1] = i;
}
else if(c[i] >1 || c[i] < -1)
return false;
}
for(int i = 1; i <= n; i++)
if(c[i]!=0)
return false;
return true;
}

void prin()
{
printf("YES\n");
printf("%d\n",tot);
for(int i = 0; i < tot; i++)
printf("%d %d\n",p[i][0],p[i][1]);
}

int main()
{
int T;
//freopen("01.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
ll sum = 0;
memset(p,0,sizeof(p));
for(int i = 1; i <= n; i++)
{
scanf("%d",&a[i]);
sum += a[i];
}

if(sum % n)
{
printf("NO\n");
continue;
}
aver = sum / n;
flag = 0;
tot = 0;
for(int j = 1; j <= n; j++)
a[j] =a[j] - aver;
c[1] = a[1];
c[2] = a[2];
if(solve())
{
prin();
}
else
{
tot = 0;
c[1] = a[1] - 1;
c[2] = a[2] + 1;
p[tot][0] = 1;
p[tot++][1] = 2;
if(solve())
prin();
else
{
tot = 0;
c[1] = a[1] + 1;
c[2] = a[2] - 1;
p[tot][0] = 2;
p[tot++][1] = 1;
if(solve())
prin();
else
printf("NO\n");
}
}

}
return 0;
}