SQYBI.com

水题,枚举即可.
第一次提交WA,没有注意这道题并不是Special Judge,如果有多组顺序符合要求,需要输出字典序最靠前的一组.
虽然用了string,不过代码写的还是很垃圾.

/*
  UVa 102; Ecological Bin Packing
  - sqybi's code
*/
//for my winsty

#include
#include
using namespace std;

const int nn = 3;
int t, res, now;
int a[nn][nn];
bool u[nn];
char ch[nn] = {'B', 'G', 'C'};
string ress, nows;

void get(int i)
{
    if (i == 3)
    {
        if (now > res || (now == res && nows < ress))
        {
            res = now;
            ress = nows;
        }
        return;
    }
    for (int j = 0; j != 3; ++ j)
        if (! u[j])
        {
            u[j] = true;
            now = now + a[i][j];
            nows += ch[j];
            get(i+1);
            nows = nows.erase(i, 1);
            now = now - a[i][j];
            u[j] = false;
        }
}

int main()
{
    while (scanf("%d", &a[0][0]) != EOF)
    {
        t = a[0][0];
        for (int i = 0; i != 3; ++ i)
            for (int j = 0; j != 3; ++ j)
                if (i + j)
                {
                    scanf("%d", &a[i][j]);
                    t += a[i][j];
                }
        res = 0;
        now = 0;
        nows = "";
        memset(u, false, sizeof(u));
        get(0);
        cout << ress << ' ' << t - res << endl;
    }
}

经典的3n+1,没有想到这道题调了一个晚上.
在Project Euler上看过这道题,于是按照那道题写了个部分DP.然后诡异的WA了.
调啊,调啊,只找到了一个trick:输入的两个数不一定小数在前大数在后.
然后呢,我这个程序是search,可加了一个预处理,for i=1~10000 do dp[i]=f(i),然后调用f(i)的地方改成dp[i],就过不了,都鬼了...怀疑数据范围有问题.

search可以1s过掉.

/*
  UVa 100; The 3n + 1 problem
  - sqybi's code
*/
//for my winsty

//#define DEBUG
#include
using namespace std;

const int nn = 1000001, mm = 10001;
int l, r, ll, rr, maxa;

int f(int i)
{
    if (i == 1)
        return 1;
    else if (i % 2)
        return f(i*3+1)+1;
    else
        return f(i/2)+1;
}

int main()
{
    #ifdef DEBUG
    freopen("uva100.in", "r", stdin);
    freopen("uva100.out", "w", stdout);
    #endif

    while (scanf("%d%d", &ll, &rr) != EOF)
    {
        l = ll;
        r = rr;
        if (l > r) swap(l, r);
        maxa = 0;
        for (int i = l; i <= r; ++ i)
            if (f(i) > maxa) maxa = f(i);
        printf("%d %d %d\n", ll, rr, maxa);
    }
}