Codeforces 740A Alyona and copybooks
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题意:某人现有n本书,想再拥有k本书,使得(n % k) == 0,求买k本书最少花多少钱。已知a元买1本书,b元买2本书,c元买3本书。
分析:n对4取余,可分为4种情况:(t = n % 4)(每个情况下考虑a,b,c的所有组合)
1、t == 0,则不必花钱买书。
2、t == 3,还需买1本,
(1)用a元买1本
(2)用c元买3本(因为a可能很大,然后c很小)
(3)肯定不能完全用b元买,因为整除不了,此外肯定不能用a + b 或 a + c买,因为这样花的钱大于a,所以就用b + c买1本
3、t == 2,还需买2本,同理,完全用a买,完全用b买,完全用c买,没必要用a + b,a + c等买
4、t == 1,还需买3本,与上面同理
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1}; const int dc[] = {-1, 1, 0, 0}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 100000 + 10; const int MAXT = 10000 + 10; using namespace std; int main(){ ll n, a, b, c; while(scanf("%lld%lld%lld%lld", &n, &a, &b, &c) == 4){ if(n % 4 == 0){ printf("0\n"); continue; } ll tmp = ll(4) - ll(n % 4); if(tmp == ll(1)){ ll t = Min(a, 3 * c); t = Min(t, b + c); printf("%lld\n", t); continue; } if(tmp == ll(2)){ ll t = Min(2 * a, b); t = Min(2 * c, t); printf("%lld\n", t); continue; } if(tmp == (3)){ ll t = Min(3 * a, c); t = Min(t, a + b); printf("%lld\n", t); continue; } } return 0; }
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