題目鏈接

Problem Description

Little Ruins is a studious boy, recently he learned the four operations!Now he want to use four operations to generate a number, he takes a string which only contains digits ‘1’ - ‘9’, and split it into 55 intervals and add the four operations ‘+’, ‘-‘, ‘*’ and ‘/’ in order, then calculate the result(/ used as integer division).Now please help him to get the largest result.

Input

First line contains an integer TT, which indicates the number of test cases.Every test contains one line with a string only contains digits ‘1’-‘9’.Limits1≤T≤1051≤T≤1055≤length of string≤205≤length of string≤20

Output

For every test case, you should output ‘Case #x: y’, where x indicates the case number and counts from 1 and y is the result.

Sample Input

112345

Sample Output

Case #1: 1

AC

• 因為固定順序，所以只有一種形式：A + B - C * D / E,貪心使得（A+ B）越大，（C* D / E)越小，結(jié)果的大小關(guān)鍵在于（A + B），所以C、D、E都只占一位數(shù)最好，但是有可能出現(xiàn)類似9 * 9 / 1的情況，所以比較兩種方法，一個是C、D、E分別占一位數(shù)，一個是C、D都分別占一位數(shù)，E占兩位

#include <iostream>#include <stdio.h>#include <map>#include <vector>#include <algorithm>#define N 100005#define ll long longusing namespace std;ll num(char c) {return c - '0';}string s;// 貪心得到和的最大值 ll get_sum(int l, int r) {if (r - l < 1) return -1e15;ll t1 = 0, t2 = 0, t3 = 0, t4 = 0;for (int i = l; i < r; ++i) {t1 = t1 * 10 + s[i] - '0';}t2 = s[r] - '0';for (int i = l + 1; i <= r; ++i) {t3 = t3 * 10 + s[i] - '0';}t4 = s[l] - '0';return max(t1 + t2, t3 + t4);}int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endifint t;scanf("%d", &t);int Case = 1;while (t--) {ll ans1, ans2;cin >> s;int len = s.size(); ans1 = get_sum(0, len - 4) - num(s[len - 3]) * num(s[len - 2]) / num(s[len - 1]);ans2 = get_sum(0, len - 5) - num(s[len - 4]) * num(s[len - 3]) / (num(s[len - 2]) * 10 + num(s[len - 1]));printf("Case #%d: %lld\n", Case++, max(ans1, ans2));}return 0;}

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