¶
하,
하
행 .
한.
ex)
43
1 3 ->
2 4
n = bq+r
43 = 14*3+1 = 3333333333333+3+1 = 3333333333333+4 => 13, 1
하
행 .
한.
ex)
43
1 3 ->
2 4
n = bq+r
43 = 14*3+1 = 3333333333333+3+1 = 3333333333333+4 => 13, 1
35
1 5 ->
20 7
35 = 7*5+0 = 5555555+0 =>7, 0
1 5 ->
20 7
35 = 7*5+0 = 5555555+0 =>7, 0
< 05 05 15>
n2 크 행 .
n2 worst case 한.
2000000000
1 3
1900000000 1900000000
하 히 하 함. 해;;
.. ㅠㅠ
n2 크 행 .
n2 worst case 한.
2000000000
1 3
1900000000 1900000000
하 히 하 함. 해;;
.. ㅠㅠ
¶
~cpp
#include <iostream>
using namespace std;
void main(){
int N[100], n1[100], c1[100], n2[100], c2[100];
int tot_arr = 0;
for(int i=0;i<100;i++){
cin >> N[i];
if(N[i] == 0)
break;
cin >> c1[i]>> n1[i]>> c2[i] >> n2[i];
tot_arr++;
}
for(i=0;i<tot_arr;i++){
if(c1[i]/(double)n1[i] > c2[i]/(double)n2[i]) //n1 함.
swap(n1[i],n2[i]);
int r = N[i]%n1[i];
int m1 = N[i]/n1[i];
int m2 = -1;
for(int k=0; k<=n2[i] ;k++){
if(r%n2[i]==0){
m2=r/n2[i];
cout <<n1[i]<<" "<<m1<<" "<<n2[i]<<" "<<m2<<" "<<"\n";
break;
}
r = ((n2[i]/n1[i]==0?1:n2[i]/n1[i])*k+1)*n1[i] + (N[i]%n1[i]); //r n2 큰 n1 N[i]%n1[i] 한 .
if(r-n2[i]>=n1[i])
r-=n1[i];
if(r>N[i] || r<0){
int tempk;
if(n2[i]*k>N[i])
tempk = k-1;
else
tempk = k;
if((N[i]-n2[i]*tempk)%n1[i] == 0)
r = n2[i]*tempk;
else
break;
}
m1=(N[i]-r)/n1[i];
}
if(m2 == -1)
cout << "failed\n";
cout <<"total " << k << " roops\n";
}
}
~cpp
#include <iostream>
using namespace std;
void main(){
int N[100], n1[100], c1[100], n2[100], c2[100];
int tot_arr = 0;
for(int i=0;i<100;i++){
cin >> N[i];
if(N[i] == 0)
break;
cin >> c1[i]>> n1[i]>> c2[i] >> n2[i];
tot_arr++;
}
for(i=0;i<tot_arr;i++){
if(c1[i]/(double)n1[i] > c2[i]/(double)n2[i]) //n1 함.
swap(n1[i],n2[i]);
int r = N[i]%n1[i];
int m1 = N[i]/n1[i];
int m2 = -1;
for(int roop=0; roop<=n2[i] ; roop++){ // n2 . r n1 n2 해 n2*n1+r
// (n2*n1+r)%n2 == r%n2 roop=n2 환 .
if(r%n2[i]==0){
m2=r/n2[i];
cout <<n1[i]<<" "<<m1<<" "<<n2[i]<<" "<<m2<<" "<<"\n";
break;
}
r+=n1[i];
m1--;
}
if(m2 == -1)
cout << "failed\n";
}
}
