Problem D: 买iphone
Time Limit: 3 Sec Memory Limit: 128 MBSubmit: 825 Solved: 121
Description
自从上次仓鼠中了1000万彩票后,彻底变成了土豪了,一群人愿意认他做干爹,仓鼠决定送他的干儿子每人一部iphone。仓鼠今天带了一群人去买iphone,每个人身上都背着一个大背包,只有3种背包,分别能装a,b,c个iphone,仓鼠希望每个人都能用iphone装满自己的背包。仓鼠有n个干儿子,也就是要买n部手机,仓鼠带了k个人去买手机,问是否存在一种情况能让每个买iphone的人的背包都装满
Input
输入包含多组测试数据,每组一行,包含5个整数n, k, a, b, c
(1 <= n, k , a, b, c <= 10000)
abc可以相同
Output
如果存在输出Yes,否则输出No
Sample Input
10 3 3 3 4 20 7 1 1 1 80 3 100 3 3 15 4 3 5 6
Sample Output
Yes No No Yes
HINT
样例一3个人分别带大小为 3 3 4的背包去买手机正好每个人都装满
样例二 很明显只有一种背包,全部都是1也无法买20部iphone
样例四 4个人分别带大小为 3 3 3 6的背包去买手机正好每个人都装满
# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <vector> # include <bitset> # include <cmath> # include <cstdlib> # define INF 0x3f3f3f3f # define LL long long using namespace std; bitset<10005>dp, pd; int main() { int n, k, a, b, c; while(~scanf("%d%d%d%d%d",&n,&k,&a,&b,&c)) { dp.reset(); dp[0] = 1; int imin = min(a, min(b, c)); int imax = max(a, max(b, c)); if(k*imin > n || k*imax < n) { puts("No"); continue; } for(int i=1; i<=k; ++i) { swap(dp, pd); dp.reset(); dp |= pd << a; dp |= pd << b; dp |= pd << c; } if(dp[n]) puts("Yes"); else puts("No"); } return 0; }