1 条题解
-
0
C++ :
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; const int MAXN = 85, Mod = 10000; //高精四位压缩大法好 int n, m; int ar[MAXN]; struct HP { int p[505], len; HP() { memset(p, 0, sizeof p); len = 0; } //这是构造函数,用于直接创建一个高精度变量 void print() { printf("%d", p[len]); for (int i = len - 1; i > 0; i--) { if (p[i] == 0) { printf("0000"); continue; } for (int k = 10; k * p[i] < Mod; k *= 10) printf("0"); printf("%d", p[i]); } } //四位压缩的输出 } f[MAXN][MAXN], base[MAXN], ans; HP operator + (const HP &a, const HP &b) { HP c; c.len = max(a.len, b.len); int x = 0; for (int i = 1; i <= c.len; i++) { c.p[i] = a.p[i] + b.p[i] + x; x = c.p[i] / Mod; c.p[i] %= Mod; } if (x > 0) c.p[++c.len] = x; return c; } //高精+高精 HP operator * (const HP &a, const int &b) { HP c; c.len = a.len; int x = 0; for (int i = 1; i <= c.len; i++) { c.p[i] = a.p[i] * b + x; x = c.p[i] / Mod; c.p[i] %= Mod; } while (x > 0) c.p[++c.len] = x % Mod, x /= Mod; return c; } //高精*单精 HP max(const HP &a, const HP &b) { if (a.len > b.len) return a; else if (a.len < b.len) return b; for (int i = a.len; i > 0; i--) if (a.p[i] > b.p[i]) return a; else if (a.p[i] < b.p[i]) return b; return a; } //比较取最大值 void BaseTwo() { base[0].p[1] = 1, base[0].len = 1; for (int i = 1; i <= m + 2; i++){ //这里是m! m! m! 我TM写成n调了n年... base[i] = base[i - 1] * 2; } } //预处理出2的幂 int main(void) { scanf("%d%d", &n, &m); BaseTwo(); while (n--) { memset(f, 0, sizeof f); for (int i = 1; i <= m; i++) scanf("%d", &ar[i]); for (int i = 1; i <= m; i++) for (int j = m; j >= i; j--) { //因为终值是小区间,DP自然就从大区间开始 f[i][j] = max(f[i][j], f[i - 1][j] + base[m - j + i - 1] * ar[i - 1]); f[i][j] = max(f[i][j], f[i][j + 1] + base[m - j + i - 1] * ar[j + 1]); } //用结构体重载运算符写起来比较自然 HP Max; for (int i = 1; i <= m; i++) Max = max(Max, f[i][i] + base[m] * ar[i]); ans = ans + Max; //记录到总答案中 } ans.print(); //输出 return 0; }
- 1
信息
- ID
- 1887
- 时间
- 1000ms
- 内存
- 125MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者