codeforces 282B - Painting Eggs

B. Painting Eggs

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 a_i and g_i each (0 ≤ a_i, g_i ≤ 1000; a_i + g_i = 1000): a_i is the price said by A. for the i-th egg and g_i 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 S_a, and the money Uncle J. must pay G. for the painting as S_g, then this inequality must hold: |S_a  -  S_g|  ≤  500.

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

Sample test(s)

input

21 999999 1

output

AG

input

3400 600400 600400 600

output

AGA

题意是这样的 大致是：让你处理n个鸡蛋 然后A和G对于每个序号鸡蛋都给出自己的价格，然后让你从A和G中选择一个来处理的鸡蛋，n个鸡蛋恰当的选择，要使最后的A和G的到的总工资的差的绝对值要小于等于500就可以了，并按顺序输出选择每个鸡蛋是让A服务还是G服务。（输出任何一个结果皆可） 若没有满足条件输出-1.

算法： 看到这道题的数据量有1e6，就想到的是n或n*logn的算法。  发现可以用n*logn的排序来解决。贪心的算法（排序两次即可）

见代码：

#include<stdio.h>#include<iostream>#include<algorithm>#include<stdlib.h>#include<math.h>using namespace std;struct node{                  //rank 记录输入数据，judge记录是这个鸡蛋归谁处理，a是A处理的工资，b是G处理的工资         int a,b,rank,judge;       }egg[1100000];int cmp(node a,node b){    return a.a<b.a;    }int cmp1(node a,node b){    return a.rank<b.rank;    }int main(){    int n;    while(scanf("%d",&n)!=EOF){        int i;        for(i=1;i<=n;i++){           scanf("%d%d",&egg[i].a,&egg[i].b);                                      egg[i].rank=i;        }        sort(egg+1,egg+n+1,cmp);//工资从大到小排序对于A来讲        int ans1=0,ans2=0,j=n;        i=1;                while(i<=j){           //key point 判断鸡蛋改归谁处理             if(ans1+egg[i].a>500){                egg[j].judge=2;                ans1-=egg[j].b;                j--;             }             else {                  ans1+=egg[i].a;                  egg[i].judge=1;                  i++;             }                 }        if(abs(ans1)>500)printf("-1");        else {                             //按输入顺序输出选择A,G的序号            sort(egg+1,egg+1+n,cmp1);            for(i=1;i<=n;i++)              if(egg[i].judge==1)printf("A");              else printf("G");        }           printf("\n");         }        return 0;}