hdu5353||2015多校联合第六场1001 贪心

http://acm.hdu.edu.cn/showproblem.php?pid=5353

Problem Description

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 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.

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

 

Input

There are multiple test cases. The first line of input contains an integer T,
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.

 

Output

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 integer m (0≤m≤n) in
the second line denoting the number of operations needed. Then each of the following m lines
contain two integers x and y (1≤x,y≤n),
which means that x-th
soda gives y-th
soda a candy.

 

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

/**
hdu5353||2015多校联合第六场1001  贪心
题目大意：给定一个序列首尾相连，每两个临的数可以相互给出一个或接受一个1，问是否可以经过一系列转移使所有的数相等
解题思路：枚举第一个数和最后一个数之间的关系（给出一个，接受一个，不处理），然后从前往后判断i和i+1的关系,i少1则从i+1得1，
多1则给i+1一个，正好则跳过。处理的时候就不用考虑i-1了，最后判断是否成功即可。复杂度O(n*3)
*/
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long LL;
int a[100005],b[100005];
int n,k,p[100005][2];

bool judge()
{
memcpy(b,a,sizeof(a));
for(int i=0;i<n-1;i++)
{
if(b[i]<0)
{
b[i]++;
b[i+1]--;
p[k][0]=i+2;
p[k++][1]=i+1;
}
else if(b[i]>0)
{
b[i]--;
b[i+1]++;
p[k][0]=i+1;
p[k++][1]=i+2;
}
}
for(int i=0;i<n;i++)
{
if(b[i]!=0)return 0;
}
return 1;
}

void print(int flag)
{
if(flag)
{
printf("YES\n%d\n",k);
for(int i=0; i<k; i++)
{
printf("%d %d\n",p[i][0],p[i][1]);
}
}
else
{
puts("NO");
}
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
LL sum=0;
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
if(sum%n)
{
puts("NO");
continue;
}
sum/=n;
if(n==2&&max(a[1],a[0])-min(a[1],a[0])>2)
{
puts("NO");
continue;
}
for(int i=0; i<n; i++)
{
a[i]-=sum;
}
k=0;
int flag=0;
if(judge())
{
flag=1;
print(1);
}
else
{
k=0;
p[k][0]=n,p[k++][1]=1;
a[n-1]--,a[0]++;
if(judge())
{
flag=1;
print(1);
}
else
{
k=0;
p[k][0]=1,p[k++][1]=n;
a[0]-=2,a[n-1]+=2;
if(judge())
{
flag=1;
print(1);
}
}
}
if(flag==0)
{
print(0);
}
}
return 0;
}

Problem Description

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 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.

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

 

Input

There are multiple test cases. The first line of input contains an integer T,
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.

 

Output

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 integer m (0≤m≤n) in
the second line denoting the number of operations needed. Then each of the following m lines
contain two integers x and y (1≤x,y≤n),
which means that x-th
soda gives y-th
soda a candy.

 

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