uva 562 Dividing coins
It’s commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus created copper-wire.
Not commonly known is that the fighting started, after the two Dutch tried to divide a bag with coins between the two of them. The contents of the bag appeared not to be equally divisible. The Dutch of the past couldn’t stand the fact that a division should favour one of them and they always wanted a fair share to the very last cent. Nowadays fighting over a single cent will not be seen anymore, but being capable of making an equal division as fair as possible is something that will remain important forever…
That’s what this whole problem is about. Not everyone is capable of seeing instantly what’s the most fair division of a bag of coins between two persons. Your help is asked to solve this problem.
Given a bag with a maximum of 100 coins, determine the most fair division between two persons. This means that the difference between the amount each person obtains should be minimised. The value of a coin varies from 1 cent to 500 cents. It’s not allowed to split a single coin.
InputA line with the number of problems n, followed by n times:a line with a non negative integer m () indicating the number of coins in the baga line with m numbers separated by one space, each number indicates the value of a coin.OutputThe output consists of n lines. Each line contains the minimal positive difference between the amount the two persons obtain when they divide the coins from the corresponding bag.Sample Input232 3 541 2 4 6Sample Output01
题目大意：每组数据给出钱的组数n，接下来是n组钱的金额数。要求把n组钱分成2份，且2份钱的金额总数相差最小，并输出相差金额数。
解题思路：将每组钱的金额数加在一起获得sum，除以二获得的h，就是01背包的容量。然后找出n组钱中的各种组合，，找出其中离h最近的，乘以二就是ans。当sum为奇数时，最后得出的ans还要加一。
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#define N 105#define M 50005typedef long long ll;using namespace std;int num[N];ll dp[M];int main() {int T;scanf("%d", &T);while (T–) {memset(dp, 0, sizeof(dp));int n, sum = 0;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &num[i]);sum += num[i];}dp[0] = 1;int h = sum / 2, ans, flag = 0;for (int i = 0; i < n; i++) {for (int j = h – 1; j >= 0; j–) {if (dp[j]) {dp[j + num[i]] = 1;}}}for (int i = h; i >= 0; i–) {if (dp[i]) {ans = (h – i) * 2;break;}}printf("%d\n", ans + (sum % 2));}return 0;}
人的一生要疯狂一次，无论是为一个人，一段情，一段旅途，或一个梦想。
