codeforces 381A. Alyona and copybooks=
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在这个题中,不能光考虑k = 0,1, 2,3. 因为条件是花钱最少,不是k越小,花钱越少。要考虑每一种k情况下的最小花钱数。例如:n % 4 == 1 的情下,k可以是3 {(1+1+1+1)%4==0(1+3)% 4 == 0 (1+1+2)%4==0 },当n % 4 == 3的情下,k可以是1 {(3+1)%4==0} ,k可以是5 {(3+2+3)% 4 == 0 },k可以是9{(3+3*3)%4 == 0}
#include<iostream>
#include<algorithm>#include
using namespace std;
using namespace std;int main(){long long n, a, b, c;while (cin >> n >> a >> b >> c){long long result;if (n % 4 == 0){result = 0;}else if (n % 4 == 1){result = min(3 * a, min(a + b, c));}else if (n % 4 == 2){result = min(2 * a, min(b, 2 * c));}else{result = min(a, min(b + c, 3 * c));}cout << result << endl;}return 0;}
int main(){
long long n, a, b, c;
while (cin >> n >> a >> b >> c){
long long result;
if (n % 4 == 0){
result = 0;
}
else if (n % 4 == 1){
result = min(3 * a, min(a + b, c));
}
else if (n % 4 == 2){
result = min(2 * a, min(b, 2 * c));
}
else{
result = min(a, min(b + c, 3 * c));
}
cout << result << endl;
}
return 0;
}
0 0
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