Expected
Run
makeChocolate(4, 1, 9) → 4
4
OK
makeChocolate(4, 1, 10) → -1
-1
OK
makeChocolate(4, 1, 7) → 2
2
OK
makeChocolate(6, 2, 7) → 2
-1
X
makeChocolate(4, 1, 5) → 0
0
OK
makeChocolate(4, 1, 4) → 4
4
OK
makeChocolate(5, 4, 9) → 4
-1
X
makeChocolate(9, 3, 18) → 3
3
OK
makeChocolate(3, 1, 9) → -1
-1
OK
makeChocolate(1, 2, 7) → -1
-1
OK
makeChocolate(1, 2, 6) → 1
-1
X
makeChocolate(1, 2, 5) → 0
-1
X
makeChocolate(6, 1, 10) → 5
0
X
makeChocolate(6, 1, 11) → 6
1
X
makeChocolate(6, 1, 12) → -1
-1
OK
makeChocolate(6, 1, 13) → -1
-1
OK
makeChocolate(6, 2, 10) → 0
0
OK
makeChocolate(6, 2, 11) → 1
1
OK
makeChocolate(6, 2, 12) → 2
2
OK
makeChocolate(60, 100, 550) → 50
50
OK
makeChocolate(1000, 1000000, 5000006) → 6
6
OK
makeChocolate(7, 1, 12) → 7
2
X
makeChocolate(7, 1, 13) → -1
-1
OK
makeChocolate(7, 2, 13) → 3
3
OK
other tests
X

public int makeChocolate(int small, int big, int goal) {
int sum=-1;
if((small+big*5)<goal){sum=-1;}
if(!(big==0)){
int bigCount=goal/5;//需要的大砖块数
int bigSum=bigCount*5;//大砖块值的总和
int yue=goal-bigSum;//需要的小砖块数
if(big<bigCount){
if(big*5+small>=goal){
sum=goal-big*5;}
}else{
if(bigSum+small>=goal){
sum=goal-bigCount*5;}
}
}else if(small>goal){
sum=goal;
}
return sum;
}