CodeForces - 282B Painting Eggs

YinJianxiang發表於2017-11-30
AI

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 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.

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

題目大意：就是有兩個序列，A和G，A[i]+G[i]為1000，每次從序列中一個取出一個元素，同時從A序列取的元素代數和sum_a和從G序列取的元素代數和sum_g之差不超過500，如果存在列印取的順序，沒有的話就輸出-1


大體想法：貪心演算法，既然使取元素兩邊之差不超過500，則可以使每一步取完之後的之差不超過500，這樣顯然是滿足條件的。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>

using namespace std;


int main() {
    int n;
    
    scanf("%d",&n);
    
    int a = 0;
    int g = 0;
    bool flag = true;
    string s = "";
    int index = 0;
    
    for(int i = 0;i < n;i++) {
        int x,y;
        
        scanf("%d %d",&x,&y);
        if(a+x-g <= 500) {
            a += x;
            s.append("A");
        } else if(g+y-a <= 500){
            g += y;
            s.append("G");
        } else {
            flag = false;
            break;
        }
    }
    if(flag) {
        cout << s << "\n";
    } else {
        printf("-1\n");
    }
    
    return 0;
}


















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