NC13882. Knapsack Problem
描述
输入描述
The first line contains an positive integer T(1≤T≤10), represents there are T test cases.For each test case:The first line contains three positive integers n,w,d(1≤n≤100,1≤w≤100,1≤d≤100) - the number of items HH wants to buy, the max weight that his bag can carry, and the money he has.The second line contains n integers w1,w2…wn(1≤wi≤100). The third line contains n integers c1,c2…cn(1≤ci≤100).
输出描述
For each test case, output one line "YES" (without quotes) if HH is possible to buy all the items and carry them in his bag, and "NO" (without quotes) otherwise.
示例1
输入:
2 4 12 17 1 2 4 5 5 4 6 2 4 11 17 1 2 4 5 5 4 6 2
输出:
YES NO
说明:
In the first example all the items cost 17 dollars in total and weight 12 kilograms in total, HH has enough money and his bag can carry 12 kilogram things.C++11(clang++ 3.9) 解法, 执行用时: 10ms, 内存消耗: 488K, 提交时间: 2020-02-11 19:41:30
#include<bits/stdc++.h>
using namespace std;
int main()
{int t,w,n,d;
cin>>t;
while(t--){
cin>>n>>d>>w;
int p=0,q=0,x,y;
for(int i=0;i<n;i++){
cin>>x;
p+=x;
}
for(int i=0;i<n;i++){
cin>>y;
q+=y;
}
if(p<=d&&q<=w) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
C(clang11) 解法, 执行用时: 3ms, 内存消耗: 368K, 提交时间: 2020-11-26 12:25:43
#include<stdio.h>
int t,n,c,w;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&w,&c);
int x;
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
w-=x;
}
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
c-=x;
}
if(w>=0&&c>=0)printf("YES\n");
else printf("NO\n");
}
return 0;
}
Python3 解法, 执行用时: 35ms, 内存消耗: 4528K, 提交时间: 2022-06-22 14:11:54
T=int(input())
for i in range(T):
c,w,d=map(int,input().split())
a=map(int,input().split())
b=map(int,input().split())
if sum(a)<=w and sum(b)<=d:
print("YES")
else:
print("NO")