ZOJ 3657 The Little Girl who Picks Mushrooms

The Little Girl who Picks Mushrooms

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on ZJU. Original ID: 3657
64-bit integer IO format: %lld      Java class name: Main

 

 

It's yet another festival season in Gensokyo. Little girl Alice planned to pick mushrooms in five mountains. She brought five bags with her and used different bags to collect mushrooms from different mountains. Each bag has a capacity of 2012 grams. Alice has finished picking mushrooms in 0 ≤ n ≤ 5 mountains. In the the i-th mountain, she picked 0 ≤wi ≤ 2012 grams of mushrooms. Now she is moving forward to the remained mountains. After finishing picking mushrooms in all the five mountains, she want to bring as much mushrooms as possible home to cook a delicious soup.

Alice lives in the forest of magic. At the entry of the forest of magic, lives three mischievous fairies, Sunny, Lunar and Star. On Alice's way back home, to enter the forest, she must give them exactly three bags of mushrooms whose total weight must be of integral kilograms. If she cannot do so, she must leave all the five bags and enter the forest with no mushrooms.

Somewhere in the forest of magic near Alice's house, lives a magician, Marisa. Marisa will steal 1 kilogram of mushrooms repeatedly from Alice until she has no more than 1 kilogram mushrooms in total.

So when Alice get home, what's the maximum possible amount of mushrooms Alice has? Remember that our common sense doesn't always hold in Gensokyo. People in Gensokyo belive that 1 kilograms is equal to 1024 grams.

Input

There are about 8192 test cases. Process to the end of file.

The first line of each test case contains an integer 0 ≤ n ≤ 5, the number of mountains where Alice has picked mushrooms. The second line contains n integers 0 ≤ wi ≤ 2012, the amount of mushrooms picked in each mountain.

Output

For each test case, output the maximum possible amount of mushrooms Alice can bring home, modulo 20121014 (this is NOT a prime).

Sample Input

1
9
4
512 512 512 512
5
100 200 300 400 500
5
208 308 508 708 1108

Sample Output

1024
1024
0
792

Note

In the second sample, if Alice doesn't pick any mushrooms from the 5-th mountain. She can give (512+512+0)=1024 grams of mushrooms to Sunny, Lunar and Star. Marisa won't steal any mushrooms from her as she has exactly 1 kilograms of mushrooms in total.

In the third sample, there are no three bags whose total weight is of integral kilograms. So Alice must leave all the five bags and enter the forest with no mushrooms.

In the last sample:

• Giving Sunny, Lunar and Star: (208+308+508)=1024
• Stolen by Marisa: ((708+1108)-1024)=792

 

Source

The 2012 ACM-ICPC Asia Changchun Regional Contest

Author

WU, Zejun

 

解题：题目不好理解。苦逼的Alice.

 

采姑娘的小蘑菇。话说有五座山，输入n，表示已经踩过几座山了。剩余的几座山还是会去采的。如果n少于3.那么任选已经采过的两座山，从未选的山里面采取一些蘑菇，使得三山的蘑菇量刚好是1024的整数倍。

 

如果n == 4.那么，如果已采的4座中，有3座山的蘑菇量和刚好是1024的整数倍。那么，剩余的已采的那座山和未采的那座山，一定可使剩余的蘑菇量为1024.

如果已采的4座中，找不到三山的蘑菇量和为1024的整数倍，那么，只好选择其中的两座山，与未采的那座山，求和，使得这三山的蘑菇量为1024.前面采过的剩余的那两座山的蘑菇量即为剩余的（当然要不断-1024，使得不多于1024）。

 

如果n == 5,那么只看五选三山中是否存在三山蘑菇量和为1024，如有，则剩余的两山蘑菇和为剩余的蘑菇。（while(x > 1024)   x -= 1024;) .否则，0.

 

以上求解，均取最优的那个。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
int d[6],n,ans,sum;
bool flag;
int main(){
    while(~scanf("%d",&n)){
        sum = ans = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d",d+i);
            sum += d[i];
        }
        if(n <= 3) {puts("1024");continue;}
        if(n == 5){
            for(int i = 1; i <= n; i++){
                for(int j = i+1; j <= n; j++){
                    for(int k = j+1; k <= n; k++){
                        int tmp  = d[i]+d[j]+d[k];
                        if(tmp%1024 == 0){
                            tmp = sum - tmp;
                            while(tmp > 1024) tmp -= 1024;
                            ans = max(ans,tmp);
                        }
                    }
                }
            }
            printf("%d
",ans);
            continue;
        }
        flag = false;
        for(int i = 1; i <= n; i++){
            for(int j = i+1; j <= n; j++){
                for(int k = j+1; k <= n; k++){
                    int tmp = d[i]+d[j]+d[k];
                    if(tmp%1024 == 0){
                        flag = true;
                        break;
                    }
                }
            }
        }
        if(flag){puts("1024");continue;}
        for(int i = 1; i <= n; i++){
            for(int j = i+1; j <= n; j++){
                int tmp = d[i]+d[j];
                while(tmp > 1024) tmp -= 1024;
                ans = max(ans,tmp);
            }
        }
        printf("%d
",ans);
    }
    return 0;
}