Codeforces Round #326 (Div. 2)

贪心 A - Duff and Meat

题意：卖肉问题，当天的价格便宜可以将以后的肉买了，问最少花费多少钱

分析：差点就做不出来了，维护一个动态的前缀最小值，如果当前的价格便宜则更新最小值，当天的肉用最小值买。

#include <cstdio>

const int N = 1e5 + 10;
int a[N], p[N];

int main(void)	{
	int n;	scanf ("%d", &n);
	int ans = 0, mn = 111;
	for (int i=1; i<=n; ++i)	{
		scanf ("%d%d", &a[i], &p[i]);
		if (p[i] < mn)	mn = p[i];
		ans += a[i] * mn;
	}
	printf ("%d
", ans);

	return 0;
}

　　

水+约数枚举 B - Duff in Love

暴力枚举一下就可以了

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

typedef long long ll;

void divisor(ll n, vector<ll> &ret)	{
	ret.clear ();
	for (ll i=1; i*i<=n; ++i)	{
		if (n % i == 0)	{
			ret.push_back (i);
			if (n / i != i)	ret.push_back (n / i);
		}
	}
	ret.push_back (n);
	sort (ret.begin (), ret.end ());
}

int main(void)	{
	ll n;	scanf ("%I64d", &n);
	if (n == 1)	{
		puts ("1");	return 0;
	}
	vector<ll> div;
	divisor (n, div);
	for (int i=div.size ()-1; i>=0; --i)	{
        ll x = div[i];
        if (sqrt (x) * sqrt (x) == x)	continue;
        bool flag = true;
        for (ll j=2; j*j<=x; ++j)	{
			ll y = j * j;
			if (x % y == 0)	{
				flag = false;	break;
			}
        }
        if (flag)	{
			printf ("%I64d
", x);	return 0;
        }
	}

	return 0;
}

　　

水 C - Duff and Weight Lifting

题意：1024游戏

分析：考虑两个两个进位，如果是奇数的多一个步骤，在最后统计要累加的步骤数

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

const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int cnt[N];
int w[N];

int main(void)	{
	int n;	scanf ("%d", &n);
	int mx = -1, mn = INF;
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &w[i]);
		if (mx < w[i])	mx = w[i];
		if (mn > w[i])	mn = w[i];
		cnt[w[i]]++;
	}
    int ans = 0, pre = 0;
    for (int i=mn; i<=mx; ++i)	{
		if (i == mn)	{
			pre = cnt[i] / 2;
			if (cnt[i] & 1)	ans++;
		}
		else	{
			cnt[i] += pre;
			pre = cnt[i] / 2;
			if (cnt[i] & 1)	ans++;
		}
    }
    while (pre)	{
		if (pre & 1)	ans++;
		pre >>= 1;
    }
    printf ("%d
", ans);

	return 0;
}