UVa 311 - Packets (构造)
题意
给出各个型号的箱子的个数,求最少要用几个6*6的装。
思路
以前看过这题,那时候觉得一种一种情况讨论太麻烦了,又想不到其他的方法,就跳过去了(掩面)
今天又看了一下,觉得思路还是挺清楚的,分类讨论。。。看来我是量变了一下了。
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <cstdlib>
#include <fstream>
#include <string>
#include <sstream>
#include <map>
#include <cmath>
#define LL long long
#define lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 4000 + 5;
const int MOD = 20071027;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int num[10];
void Add()
{
if (num[2] < 0)
{
num[1] -= abs(num[2]) * 4;
num[2] = 0;
}
}
void Solve(int n, int m)
{
num[2] -= n;
num[1] -= m;
Add();
}
int main()
{
//ROP;
int i, j;
while (~scanf("%d", #[1]))
{
int cnt = 0, temp = 0;
for (i = 2; i < 7; i++) scanf("%d", #[i]);
for (i = 1; i < 7; i++) temp |= num[i];
if (!temp) break;
cnt += num[6];
cnt += num[5];
num[1] -= 11 * num[5];
if (num[4])
{
cnt += num[4];
num[2] -= num[4] * 5;
Add();
}
if (num[3])
{
int mod = num[3] % 4;
cnt += (num[3] >> 2);
if (mod)
{
cnt++;
if (mod == 1) Solve(5, 7);
else if (mod == 2) Solve(3, 6);
else if (mod == 3) Solve(1, 5);
}
}
if (num[2] > 0)
{
cnt += num[2] / 9;
int mod = num[2] % 9;
if (mod) cnt++;
num[1] -= 36 - mod * 4;
}
if (num[1] > 0)
{
cnt += num[1] / 36;
if (num[1] % 36) cnt++;
}
printf("%d\n", cnt);
}
return 0;
}