Codeforces 214b hometask

In this question, a large number composed of N numbers must be a multiple of 2, 3, and 5.

Because it is a multiple of 2 and 5, the single digit is 0. If n is not 0,-1 is output directly;

The difficulty is that it must be a multiple of 3.

If a number is a multiple of three, the sum of all its numbers must be a multiple of three.

When the sum of N numbers is a multiple of 3, it can be output from large to small;

When the sum of N numbers is not a multiple of 3, if the number of N has the same number as the remainder of mod 3 and the remainder of mod 3, remove the smallest of them, otherwise, the remainder of the two modulo 3 values is not 0, and the remainder of the two modulo 3 values is the smallest. If there is no such number, output-1;

Note that when the result is 0, leading 0 cannot be output;

 

PS: I started to write a very complex code. Later I read the 60 ms algorithm of the cool guys and then I typed it again... The result is 92 ms. Kneeling orz for Daniel ~~ O (>_<) O ~~

Appendix: (the second time is a little concise code)

# Include <iostream>
# Include <cstring>
Using namespace STD;

Int main (){
Int N;
Int sign [5];
Int num [20];
Int sum;
While (CIN> N ){
Sum = 0;
Sign [0] = sign [1] = sign [2] = 100;
Memset (Num, 0, sizeof num );
For (INT I = 0; I <n; I ++ ){
Int X;
Cin> X;
Num [x] ++;
Sum + = X;
Sign [x % 3] = min (sign [x % 3], X );
}
If (Num [0] = 0) {// n does not contain 0;
Cout <-1 <Endl;
Continue;
}
If (sum % 3 ){
If (sign [Sum % 3] = 100 ){
Int temp = 2;
For (INT I = 0; I <10; I ++) {// cout <I <"" <num [I] <Endl;
If (I % 3! = 0 & num [I]) {
Num [I] --;
Temp --;
N --; // prepare for removing leading zeros from the end;
}
If (I % 3! = 0 & num [I]) {
Num [I] --;
Temp --;
N --;
}
If (temp = 0)
Break;
}
If (temp ){
Cout <-1 <Endl;
Continue;
}
}
Else num [sign [Sum % 3] --, n --;
}
If (Num [0] = N) {// if the result is 0, leading zero cannot be output;
Cout <0 <Endl;
Continue;
}
For (INT I = 9; I> = 0; I --)
While (Num [I] --)
Cout <I;
Cout <Endl;
}
Return 0;
}

 

 

 

Appendix: (complex code for the first time)

# Include <iostream>
# Include <cstring>
# Include <algorithm>

# Define maxn 100000 + 10

Using namespace STD;

Int sum, ans;
Int N;
Int sign [5];
Int A [maxn];

Int main (){
While (CIN> N ){
Sum = 0;
Sign [0] = sign [1] = sign [2] = 100;
For (INT I = 0; I <n; I ++ ){
Cin> A [I];
Sum + = A [I];
Int temp;
Temp = A [I] % 3;
If (temp ){
Sign [temp] = min (sign [temp], a [I]);
}
}
Sort (A, A + n );
If (A [0]! = 0 ){
Cout <"-1" <Endl;
Continue;
}
If (sum % 3 = 0 ){
Int flag = 0;
For (INT I = n-1; I> = 0; I --){
If (flag = 0 & A [I] = 0)
Break;
Cout <A [I];
Flag = 1;
}
If (flag = 0)
Cout <"0 ";
Cout <Endl;
Continue;
}
Else if (sum % 3 = 1 ){
If (sign [1] = 100 ){
Int temp = 2;
For (INT I = 0; I <n; I ++ ){
If (A [I] % 3 = 2 ){
A [I] = 100;
Temp --;
}
If (temp = 0)
Break;
}
If (temp ){
Cout <"-1" <Endl;
Continue;
}
Int flag = 0;
For (INT I = n-1; I> = 0; I --){
If (A [I] = 100)
Continue;
If (flag = 0 & A [I] = 0)
Break;
Cout <A [I];
Flag = 1;
}
If (flag = 0)
Cout <"0 ";
Cout <Endl;
}
Else {
Int temp = 1;
Int flag = 0;
For (INT I = n-1; I> = 0; I --){
If (A [I] = sign [1] & temp ){
Temp = 0;
Continue;
}
If (flag = 0 & A [I] = 0)
Break;
Cout <A [I];
Flag = 1;
}
If (flag = 0)
Cout <"0 ";
Cout <Endl;
}
}
Else if (sum % 3 = 2 ){
If (sign [2] = 100 ){
Int temp = 2;
For (INT I = 0; I <n; I ++ ){
If (A [I] % 3 = 1 ){
A [I] = 100;
Temp --;
}
If (temp = 0)
Break;
}
If (temp ){
Cout <"-1" <Endl;
Continue;
}
Int flag = 0;
For (INT I = n-1; I> = 0; I --){
If (A [I] = 100)
Continue;
If (flag = 0 & A [I] = 0)
Break;
Cout <A [I];
Flag = 1;
}
If (flag = 0)
Cout <"0 ";
Cout <Endl;
}
Else {
Int temp = 1;
Int flag = 0;
For (INT I = n-1; I> = 0; I --){
If (A [I] = sign [2] & temp ){
Temp = 0;
Continue;
}
If (flag = 0 & A [I] = 0)
Break;
Flag = 1;
Cout <A [I];
}
If (flag = 0)
Cout <"0 ";
Cout <Endl;
}
}
}
Return 0;
}