Physical Address
304 North Cardinal St.
Dorchester Center, MA 02124
Recently, Devendra went to Goa with his friends. Devendra is well known for not following a budget.
He had Rs. ZZ at the start of the trip and has already spent Rs. YY on the trip. There are three kinds of water sports one can enjoy, with prices Rs. AA, BB, and CC. He wants to try each sport at least once.
If he can try all of them at least once output YES
, otherwise output NO
.
For each test case, print a single line containing the answer to that test case — YES
if Devendra can try each ride at least once, and NO
otherwise.
You may print each character of the string in uppercase or lowercase (for example, the strings “yEs”, “yes”, “Yes” and “YES” will all be treated as identical).
3
25000 22000 1000 1500 700
10000 7000 100 300 500
15000 5000 2000 5000 3000
NO
YES
YES
Test Case 1: After spending Rs. 2200022000 on the trip already, Devendra is left with Rs. 30003000. The water sports cost a total of 1000+1500+700=32001000+1500+700=3200. So, he does not have enough money left to try all the watersports.
Test Case 2: After spending Rs. 1000010000 on the trip, he is left with Rs. 30003000. The water sports cost a total of 100+300+500=900100+300+500=900. So, he has enough money to try all the watersports.
Test Case 3: After spending Rs. 50005000 on the trip, he is left with Rs. 1000010000. The water sports cost a total of 2000+5000+3000=100002000+5000+3000=10000. So, he still has enough money left to try all the watersports.
#include <iostream>
using namespace std;
int main() {
int T,Z,Y,A,B,C;
int cost;
cin>>T;
int arr[T][6];
for(int x=0;x<T;x++)
{
for(int y=0;y<5;y++)
{
cin>>arr[x][y];
}
}
for(int z=0;z<T;z++){
int amt_left=(arr[z][0]-arr[z][1]);
cost=(arr[z][2]+arr[z][3]+arr[z][4]);
if(amt_left>=cost){
cout<<"YES"<<endl;
}
else{
cout<<"NO"<<endl;
}
}
return 0;
}
for i in range(int(input())):
z,y,a,b,c=map(int,input().split())
if((z-y-a-b-c)>=0):
print("yes")
else:
print("no")
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