Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem F: Tug of War

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

Sample Input

1

3
100
90
200

Sample Output

190 200

---

這題比較特殊點，要求左右兩邊人數差距小於等於一。
因此原本是要會這麼寫的 dp[重量][人數] = 0 or 1
那麼因為人數至多 50，轉移一下方程將資料壓成一個 long long(64 bits)。
dp[j] |= dp[j-w]<<1;

#include <stdio.h>

int main() {
    int t, n, i, j;
    int A[100];
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int sum = 0;
        for(i = 0; i < n; i++)
            scanf("%d", &A[i]), sum += A[i];
        int hsum = sum/2, hn = n/2;
        long long dp[hsum+1];
        for(i = 0; i <= hsum; i++)
            dp[i] = 0;
        dp[0] = 1;
        for(i = 0; i < n; i++) {
            for(j = hsum; j >= A[i]; j--)
                dp[j] |= dp[j-A[i]]<<1LL;
        }
        if(n%2)
            while(!(dp[hsum]&(1LL<<hn)) && !(dp[hsum]&(1LL<<(hn+1))))
                hsum--;
        else
            while(!(dp[hsum]&(1LL<<hn)))
                hsum--;

        printf("%d %d\n", hsum, sum-hsum);
        if(t)
            puts("");
    }
    return 0;
}