感觉这个题挺有意思的, 我们可以将 L = lcm(1, 2, 3, ... , 8) 看作一组。
然后用dp[ i ][ j ]表示到第 i 种物品当前的值为 j 能用L的最大数量。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 2e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); LL W, c[9]; LL dp[10][840 * 8 + 10]; int main() { scanf("%lld", &W); for(int i = 1; i <= 8; i++) scanf("%lld", &c[i]); memset(dp, -1, sizeof(dp)); dp[0][0] = 0; for(int i = 0; i <= 7; i++) { for(int j = 0; j <= 840 * 8; j++) { if(~dp[i][j]) { LL up = 840 / (i + 1); up = min(up, c[i + 1]); for(int k = 0; k <= up; k++) { dp[i + 1][j + k * (i + 1)] = max(dp[i + 1][j + k * (i + 1)], dp[i][j] + (c[i + 1] - k) / (840 / (i + 1))); } } } } LL ans = 0; for(int i = 0; i <= 840 * 8; i++) { if(i <= W && ~dp[8][i]) { ans = max(ans, i + 840 * (min(dp[8][i], (W - i) / 840))); } } printf("%lld\n", ans); return 0; } /* */
原文:https://www.cnblogs.com/CJLHY/p/10518811.html