Painting Eggs CodeForces

The Bitlandians are quite weird people. They have very peculiar customs.

As is customary, Uncle J. wants to have n eggs painted for Bitruz (an ancient Bitland festival). He has asked G. and A. to do the work.

The kids are excited because just as is customary, they’re going to be paid for the job!

Overall uncle J. has got n eggs. G. named his price for painting each egg. Similarly, A. named his price for painting each egg. It turns out that for each egg the sum of the money both A. and G. want for the painting equals 1000.

Uncle J. wants to distribute the eggs between the children so as to give each egg to exactly one child. Also, Uncle J. wants the total money paid to A. to be different from the total money paid to G. by no more than 500.

Help Uncle J. Find the required distribution of eggs or otherwise say that distributing the eggs in the required manner is impossible.

Input
The first line contains integer n (1 ≤ n ≤ 106) — the number of eggs.

Next n lines contain two integers ai and gi each (0 ≤ ai, gi ≤ 1000; ai + gi = 1000): ai is the price said by A. for the i-th egg and gi is the price said by G. for the i-th egg.

Output
If it is impossible to assign the painting, print “-1” (without quotes).

Otherwise print a string, consisting of n letters “G” and “A”. The i-th letter of this string should represent the child who will get the i-th egg in the required distribution. Letter “A” represents A. and letter “G” represents G. If we denote the money Uncle J. must pay A. for the painting as Sa, and the money Uncle J. must pay G. for the painting as Sg, then this inequality must hold: |Sa  -  Sg|  ≤  500.

If there are several solutions, you are allowed to print any of them.

Example
Input
2
1 999
999 1
Output
AG
Input
3
400 600
400 600
400 600
Output
AGA

大致题意：有n个鸡蛋，分给A和G两个人，A对第i个鸡蛋出价ai，G对第i个鸡蛋出价bi，ai+bi=1000，如何分配能保证|Sa - Sb|<=500。

思路：因为ai+bi=1000,所以我们可以将ai转化为1000-bi，假设选择了x个鸡蛋价格按A来算，那么Sa-Sb=1000*x-b1-b2-b3-…bn，所以我们只需求出是否存在x满足|1000*x-sumb|<=500成立即可。

代码如下

#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define LL long long using namespace std; const int N=1e6+5;int f[N];int main(){    int n;    scanf("%d",&n);    int sum=0;    int x,y;    for(int i=1;i<=n;i++)    {        scanf("%d%d",&x,&y);        sum+=y;    }    if(n==1)    {        if(x<=500)        printf("A");        else         printf("G");        return 0;    }    sum+=500;    int ans=sum/1000;    int f=0;    sum-=500;    if(abs(ans*1000-sum)<=500)        f=1;    else if(abs((ans++)*1000-sum)<=500)        f=1;    if(f==0)    printf("-1");    else     for(int i=1;i<=ans;i++)    printf("A");    for(int i=ans+1;i<=n;i++)    printf("G");    return 0; }