題意

題目鏈接

Sol

看了status里面最短的代碼。。感覺自己真是菜的一批。。直接爆搜居然可以過？。。但是現在還沒終測所以可能會fst。。

#include<bits/stdc++.h> #define Pair pair<int, int> #define MP(x, y) make_pair(x, y) #define fi first #define se second #define int long long #define LL long long #define Fin(x) {freopen(#x".in","r",stdin);} #define Fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10; const double eps = 1e-9; template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;} template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;} template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;} template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;} template <typename A> inline void debug(A a){cout << a << '\n';} template <typename A> inline LL sqr(A x){return 1ll * x * x;} inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f; } int W, a[9], ans; void dfs(int x, int now) {if(x == 9) {chmax(ans, now);return ;}for(int num = 9, v = min((W - now) / x, a[x]); num; num--, v--)dfs(x + 1, now + max((int)0, v * x)); } signed main() {W = read();for(int i = 1; i <= 8; i++) a[i] = read();dfs(1, 0);cout << ans;return 0; }

轉載于:https://www.cnblogs.com/zwfymqz/p/10481631.html

總結

以上是生活随笔為你收集整理的cf1132E. Knapsack(搜索)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。