N个数，代表得到2^(i-1)次幂的花费，求构造L的最小花费
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #define INF 0x3f3f3f3f 6 using namespace std; 7 typedef long long LL; 8 9 LL val[40]; 10 bool vis[40]; 11 int N; 12 LL L; 13 14 int main() { 15 scanf("%d%lld", &N, &L); 16 val[0] = 1000000000; 17 LL tmp; 18 for (int i = 1; i <= N; i++) { 19 scanf("%lld", &tmp); 20 val[i]= min(tmp, 2 * val[i - 1]); 21 } 22 for (int i = N + 1; i <= 31; i++) { 23 val[i] = 2 * val[i - 1]; 24 } 25 int high = 0; 26 LL ans = 0; 27 for (int i = 30; i >= 0; i--) {//power 28 LL tmp = 1LL << i; 29 if (L >= tmp) { 30 L -= tmp; 31 vis[i] = 1; 32 } 33 } 34 for (int i = 0; i <= 30; i++) { 35 ans = min(ans, val[i + 1]); 36 if (vis[i]) { 37 ans += val[i + 1]; 38 } 39 } 40 printf("%lld ", ans); 41 return 0; 42 }