题意
有分别价值为1,2,3,4,5,6的6种物品，输入6个数字，表示相应价值的物品的数量，问一下能不能将物品分成两份，是两份的总价值相等，其中一个物品不能切开，只能分给其中的某一方，当输入六个0是（即没有物品了），这程序结束，总物品的总个数不超过20000
思路
裸的多重可行性背包，设dp[i]表示容量为i是否可装。状态设计看代码吧。
代码
[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, end) for (int i = begin; i <= end; i ++)
using namespace std;
int num[7];
bool dp[130000];
void zero_one_pack(int num, int V){
for (int i = V; i >= num; i --){
if (dp[i]) continue;
if (dp[i-num])
dp[i] = true;
}
return ;
}
void complete_pack(int num, int V){
for (int i = num; i <= V; i ++){
if (dp[i]) continue;
if (dp[i-num])
dp[i] = true;
}
return ;
}
void multi_pack(int num, int amount, int V){
if (num * amount >= V){
complete_pack(num, V);
}
int k = 1;
while(amount > k){
zero_one_pack(k*num, V);
amount -= k;
k <<= 1;
}
zero_one_pack(amount*num, V);
}
int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int t = 1;
while(scanf("%d %d %d %d %d %d", &num[1], &num[2], &num[3], &num[4], &num[5], &num[6])){
if (num[1] + num[2] + num[3] + num[4] + num[5] + num[6] == 0){
break;
}
printf("Collection #%d:
", t);
int sum = 0;
for (int i = 1; i <= 6; i ++){
sum += num[i] * i;
}
if (sum % 2 != 0){
puts("Can't be divided.");
puts("");
t ++;
continue;
}
sum /= 2;
MEM(dp, false);
dp[0] = true;
for (int i = 1; i <= 6; i ++){
multi_pack(i, num[i], sum);
}
if (dp[sum]){
puts("Can be divided.");
}
else{
puts("Can't be divided.");
}
puts("");
t ++;
}
return 0;
}
[/cpp]