HDOJ 5353 Average 模拟
2015-08-08 21:50
363 查看
各种情况特判,然后枚举前两个点之间的关系
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1723 Accepted Submission(s): 438
Special Judge
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
Source
2015 Multi-University Training Contest 6
/* ***********************************************
Author :CKboss
Created Time :2015年08月08日 星期六 09时35分46秒
File Name :HDOJ5353.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long int LL;
const int maxn=100100;
int n;
int a[maxn],b[maxn],f[maxn];
LL s;
void print()
{
puts("YES");
int cnt=0;
for(int i=0;i<n;i++)
if(f[i]) cnt++;
printf("%d\n",cnt);
for(int i=0;i<n;i++)
{
int nx=i+1;
if(nx==n) nx=0;
if(f[i]==1)
{
printf("%d %d\n",i+1,nx+1);
}
else if(f[i]==-1)
{
printf("%d %d\n",nx+1,i+1);
}
}
}
bool func(int x)
{
memcpy(b,a,sizeof(a));
memset(f,0,sizeof(f));
f[0]=x;
b[0]-=x; b[1]+=x;
b
=b[0];
for(int i=1;i<n;i++)
{
if(abs(b[i])>=2) return false;
if(b[i]==1)
{
f[i]=1;
b[i]-=1;
b[i+1]+=1;
}
else if(b[i]==-1)
{
f[i]=-1;
b[i]+=1;
b[i+1]-=1;
}
}
return true;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d",&n);
s=0;
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
s+=a[i];
}
if(s%n!=0) { puts("NO"); continue; }
s=s/n;
bool fg=true;
bool zero=true;
for(int i=0;i<n&&fg;i++)
{
a[i]-=s;
if(a[i]!=0) zero=false;
if(!(a[i]>=-2&&a[i]<=2)) fg=false;
}
if(fg==false) { puts("NO"); continue; }
if(zero==true) { puts("YES\n0"); continue; }
if(func(0)) print();
else if(func(1)) print();
else if(func(-1)) print();
else puts("NO");
}
return 0;
}
Average
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1723 Accepted Submission(s): 438
Special Judge
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
Source
2015 Multi-University Training Contest 6
/* ***********************************************
Author :CKboss
Created Time :2015年08月08日 星期六 09时35分46秒
File Name :HDOJ5353.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long int LL;
const int maxn=100100;
int n;
int a[maxn],b[maxn],f[maxn];
LL s;
void print()
{
puts("YES");
int cnt=0;
for(int i=0;i<n;i++)
if(f[i]) cnt++;
printf("%d\n",cnt);
for(int i=0;i<n;i++)
{
int nx=i+1;
if(nx==n) nx=0;
if(f[i]==1)
{
printf("%d %d\n",i+1,nx+1);
}
else if(f[i]==-1)
{
printf("%d %d\n",nx+1,i+1);
}
}
}
bool func(int x)
{
memcpy(b,a,sizeof(a));
memset(f,0,sizeof(f));
f[0]=x;
b[0]-=x; b[1]+=x;
b
=b[0];
for(int i=1;i<n;i++)
{
if(abs(b[i])>=2) return false;
if(b[i]==1)
{
f[i]=1;
b[i]-=1;
b[i+1]+=1;
}
else if(b[i]==-1)
{
f[i]=-1;
b[i]+=1;
b[i+1]-=1;
}
}
return true;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d",&n);
s=0;
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
s+=a[i];
}
if(s%n!=0) { puts("NO"); continue; }
s=s/n;
bool fg=true;
bool zero=true;
for(int i=0;i<n&&fg;i++)
{
a[i]-=s;
if(a[i]!=0) zero=false;
if(!(a[i]>=-2&&a[i]<=2)) fg=false;
}
if(fg==false) { puts("NO"); continue; }
if(zero==true) { puts("YES\n0"); continue; }
if(func(0)) print();
else if(func(1)) print();
else if(func(-1)) print();
else puts("NO");
}
return 0;
}
相关文章推荐
- [POJ 2444] Partition a Matrix 暴力
- selection sort(选择排序)
- 当 MUST_CHANGE 为 ON (开)时,不能将 CHECK_POLICY 和 CHECK_EXPIRATION 选项设为 OFF (关)
- Android--编程方法之外的常识
- Android中的ViewPager和 PagerAdapter的初步理解和使用
- Installing Tomcat 8 on OS X 10.10 Yosemite
- fwrite()函数和fread()函数
- Linux多线程【转载】
- codeforces Gym 100286H Hell on the Markets
- or1200于IMMU分析
- 贪心算法之Entropy
- 01 贪吃蛇
- Android View(一)
- The mook jong
- java泛型泛型类
- 通达OA表单动态加载下拉菜单数据项
- git 拉去远程分支到本地
- AndroidAnnotations使用
- scala第79讲:单例深入讲解及单例背后的链式表达式
- 大犇心得