ACM-ICPC2018焦作网络赛 Mathematical Curse（dp）

最新推荐文章于 2021-02-16 22:25:23 发布

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最新推荐文章于 2021-02-16 22:25:23 发布

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原文链接：http://www.cnblogs.com/yzm10/p/9651687.html

一位因年轻时轻视数学而被囚禁的王子，决定利用成年后的数学知识逃离城堡。面对一系列数学诅咒和房间里的巫师，王子必须在消除诅咒的同时最大化自己的怨恨值，以确保成功逃脱。

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Mathematical Curse

22.25%
1000ms
65536K

A prince of the Science Continent was imprisoned in a castle because of his contempt for mathematics when he was young, and was entangled in some mathematical curses. He studied hard until he reached adulthood and decided to use his knowledge to escape the castle.

There are $i^{th}ith room, there is a wizard who has a resentment value of a[i]a[i]. The prince has MM curses, the j^{th}jth curse is f[j]f[j], and f[j]f[j] represents one of the four arithmetic operations, namely addition('+'), subtraction('-'), multiplication('*'), and integer division('/'). The prince's initial resentment value is KK. Entering a room and fighting with the wizard will eliminate a curse, but the prince's resentment value will become the result of the arithmetic operation f[j]f[j] with the wizard's resentment value. That is, if the prince eliminates the j^{th}jth curse in the i^{th}ith room, then his resentment value will change from xx to (x\ f[j]\ a[i]x f[j] a[i]), for example, when x=1, a[i]=2, f[j]=x=1,a[i]=2,f[j]='+', then xx will become 1+2=31+2=3.$

Before the prince escapes from the castle, he must eliminate all the curses. He must go from $N\ge MN≥M). What is the maximum resentment value that the prince may have when he leaves the castle?$

Input

The first line contains an integer $\le T \le 1000)T(1≤T≤1000), which is the number of test cases.$

For each test case, the first line contains three non-zero integers: $\le N \le 1000), M(1 \le M \le 5)N(1≤N≤1000),M(1≤M≤5) and K(-1000 \le K \le 1000K(−1000≤K≤1000), the second line contains NN non-zero integers: a[1], a[2], ..., a[N](-1000 \le a[i] \le 1000)a[1],a[2],...,a[N](−1000≤a[i]≤1000), and the third line contains MM characters: f[1], f[2], ..., f[M](f[j] =f[1],f[2],...,f[M](f[j]='+','-','*','/', with no spaces in between.$

Output

For each test case, output one line containing a single integer.

样例输入复制

样例输出复制

2
6
3

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int kMaxN = 1005;
const int kMaxM = 6;

long long f[kMaxN][kMaxM];
long long g[kMaxN][kMaxM];
int a[kMaxN];
char curse[kMaxM];

int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    memset(f, 0xa0, sizeof(f));
    memset(g, 0x70, sizeof(g));
    for (int i = 0; i <= n; ++i) {
      f[i][0] = k;
      g[i][0] = k;
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    scanf("%s", curse);
    for (int j = 1; j <= m; ++j) {
      for (int i = j; i <= n; ++i) {
        f[i][j] = f[i - 1][j];
        g[i][j] = g[i - 1][j];
        if (curse[j - 1] == '+') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] + a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] + a[i]);
        } else if (curse[j - 1] == '-') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] - a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] - a[i]);
        } else if (curse[j - 1] == '*') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] * a[i]);
          f[i][j] = max(f[i][j], g[i - 1][j - 1] * a[i]);
          g[i][j] = min(g[i][j], f[i - 1][j - 1] * a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] * a[i]);
        } else if (curse[j - 1] == '/') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] / a[i]);
          f[i][j] = max(f[i][j], g[i - 1][j - 1] / a[i]);
          g[i][j] = min(g[i][j], f[i - 1][j - 1] / a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] / a[i]);
        }
        //printf("[%d][%d] = %lld %lld\n", i, j, f[i][j], g[i][j]);
      }
    }
    printf("%lld\n", f[n][m]);
  }
  return 0;
}