/* 用f[i][k]表示考慮到第i個(gè)雕塑，分成k組，可不可行（這是一個(gè)bool類型的數(shù)組） 轉(zhuǎn)移： f[i][k]=f[j][k-1],sum[i]-sum[j]合法 */ #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> using namespace std;typedef long long ll; const int max_n = 2010; const ll inf = 1e15;inline int getnum() {int ans = 0; char c; bool flag = false;while (!isdigit(c = getchar()) && c != '-');if (c == '-') flag = true; else ans = c - '0';while (isdigit(c = getchar())) ans = ans * 10 + c - '0';return ans * (flag ? -1 : 1); } int a[max_n], limit_A, limit_B, n;;namespace part1 {const int max_n_small = 110;bool able[max_n_small][max_n_small];inline bool check(ll ba, ll tar) {memset(able, 0, sizeof(able));able[0][0] = true;for (int i = 1; i <= n; i++)for (int j = 1; j <= limit_B; j++) {ll sum = a[i];for (int k = i - 1; k >= 0; k--) {if (able[k][j - 1] && (sum | ba) < tar) {able[i][j] = true;break;}sum += a[k];}}for (int i = limit_A; i <= limit_B; i++)if (able[n][i])return true;return false;}inline void solve() {ll sum = 0;for (int i = 1; i <= n; i ++)sum += a[i];sum <<= 1;int max_bit = 0;for (; sum >> max_bit; max_bit++);max_bit--;ll ans = 0;for (int i = max_bit; i >= 0; i--) {ll tar = ans | (1LL << i);if (!check(ans, tar))ans += (1LL << i);}cout << ans << endl;} }namespace part2 {const int max_n_small = 2010;bool able[max_n_small];int f[max_n_small];inline bool check(ll ba, ll tar) {memset(able, 0, sizeof(able));memset(f, 0x7f, sizeof(f));able[0] = true;f[0] = 0;for (int i = 1; i <= n; i++) {ll sum = a[i];for (int k = i - 1; k >= 0; k--) {if (able[k] && (sum | ba) < tar) {able[i] = true;f[i] = min(f[i], f[k] + 1);}sum += a[k];}}if (able[n] && f[n] <= limit_B)return true;elsereturn false;}inline void solve() {ll sum = 0;for (int i = 1; i <= n; i ++)sum += a[i];sum <<= 1;int max_bit = 0;for (; sum >> max_bit; max_bit++);max_bit--;ll ans = 0;for (int i = max_bit; i >= 0; i--) {ll tar = ans | (1LL << i);if (!check(ans, tar))ans += (1LL << i);}cout << ans << endl;} }int main() {freopen("sculpture.in", "r", stdin);freopen("sculpture.out", "w", stdout);n = getnum(); limit_A = getnum(); limit_B = getnum();for (int i = 1; i <= n; i++) a[i] = getnum();if (n > 100)part2::solve();elsepart1::solve(); }

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轉(zhuǎn)載于:https://www.cnblogs.com/hyfer/p/5904424.html

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