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UVA 714 Copying Books

Fucker — Tue, 19 Oct 2010 13:14:00 GMT

二分+贪心�Q�这�U�题目一直都是自��q��软肋�?/div>

此题��是一个最大值最��化的问题，诸如此类的什么最大最��，最��最大之�cȝ��首先惛_��的应该是二分�?/div>

二分��|��然后用贪心进行划分，在划分的时候需要注意分成k份，同时满��前面的尽可能的少。思�\��单，实现��h��有点�ȝ��(ch��)�?/div>

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

#define N 505

long long a[N];
int mk[N];

inline long long MAX(long long a, long long b)
{
    return a > b ? a : b;
}

long long Judge(long long mid, const int m, const int k)
{
    long long sum = 0, maxt = 0;
    for(int i = m - 1, j = 0; i >= 0; i--)
    {
        if(j < k - 1 && (i < k - j - 1 || sum + a[i] > mid))
        {
            maxt = MAX(sum, maxt);
            sum = a[i];
            mk[i] = 1;
            j++;
        }
        else sum += a[i];
    }
    return MAX(maxt, sum);
}

int main()
{
    int t, m, k;
    long long left, right, mid, ans;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &m, &k);
        left = right = 0;
        for(int i = 0; i < m; i++)
        {
            scanf("%lld", &a[i]);
            right += a[i];
        }
        while(left < right)
        {
            mid = (left + right) >> 1;
            ans = Judge(mid, m, k);
            if(ans > mid) left = mid + 1;
            else right = mid;
        }
        memset(mk, 0, sizeof(mk));
        Judge(left, m, k);
        printf("%lld", a[0]);
        if(mk[0]) printf(" /");
        for(int i = 1; i < m; i++)
        {
            printf(" %lld", a[i]);
            if(mk[i]) printf(" /");
        }
        printf("\n");
    }
    return 0;
}

Fucker 2010-10-19 21:14 发表评论

UVA 668 Parliament

Fucker — Sun, 26 Sep 2010 10:34:00 GMT

�q�是老贺猛啊�Q�我看到了我们学校夺金的希望。这题问了他之后他给��Z��我证明�?/div>

题意�Q�给定你一个数N�Q�要求拆分成一�p�d��C��和，要求�q�些��C��U�越大越好�?/div>

解法�Q�贪心。显然数被分成越多，它们之积��p��大，�q�个不证明了。这��h��们从2开始拆分n�Q�一直到k�Q��o(h��)Sum = 2+3+……+k�Q?/div>

则有n-sum<=k,�q�样我们�q�得把剩下的n-sum个数分配出去�Q�怎么分配呢？因�ؓ(f��)不能存在相同的数�Q�所以从后往前分配，对k,……�Q?,2依次�?,

因�ؓ(f��)前面只有k-1个数�Q�而剩余的可能有k个，�q�样前面每个数分配都�?之后�q�有多，怎么办，接着k开始加�Q�这样就O(ji��n)K了。数��^均，�U�就��大�?/div>

#include <stdio.h>

#define N 105

int a[N];

int main()
{
    int t, n, top;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        top = 0;
        for(int i = 2; n >= i; i++)
        {
            a[++top] = i;
            n -= i;
        }
        for(int i = top; n && i; i--)
        {
            a[i]++;
            n--;
        }
        a[top] += n;
        for(int i = 1; i < top; i++)
        {
            printf("%d ", a[i]);
        }
        printf("%d\n", a[top]);
        if(t) printf("\n");
    }
    return 0;
}

Fucker 2010-09-26 18:34 发表评论

UVA 410 Station Balance

Fucker — Mon, 20 Sep 2010 07:59:00 GMT

看了解题报告�Q�才知道可以贪心�Q�需要证明，具体�?http://www.algorithmist.com/index.php/UVa_410)

因�ؓ(f��)�_�ֺ�问题而一直错�Q�所以干脆都乘一个c�Q�最后再除，�q�样大大减少了误差�?/div>

#include <stdio.h>
#include <string.h>
#include <algorithm>

int a[20];

inline int abs(int x)
{
    if(x > 0) return x;
    return -x;
}

int main()
{
//    freopen("in", "r", stdin);
    int c, s, n, cas = 1;
    while(~scanf("%d %d", &c, &s))
    {
        memset(a, 0, sizeof(a));
        n = (c << 1) - s;
        int ave = 0;
        for(int i = n; i < (c << 1); i++)
        {
            scanf("%d", &a[i]);
            ave += a[i];
        }
    //    printf("ave = %lf\n", ave);
        std::sort(a, a + (c << 1));
        printf("Set #%d\n", cas++);
        double sum = 0;
        for(int i = 0; i < c; i++)
        {
            printf(" %d:", i);
            sum += abs(c * a[i] + c * a[(c << 1) - i - 1] - ave);
            if(a[i]) printf(" %d", a[i]);
            if(a[(c << 1) - i - 1]) printf(" %d", a[(c << 1) - i - 1]);
            printf("\n");
        }
        printf("IMBALANCE = %.5lf\n\n", sum / c);
    }
    return 0;
}

Fucker 2010-09-20 15:59 发表评论

UVA 311 Packets

Fucker — Wed, 15 Sep 2010 08:19:00 GMT

看别人思�\和代码写的�?/div>

#include <stdio.h>

int a[10];
int one[4] = {0, 7, 6, 5};
int two[4] = {0, 5, 3, 1};

int main()
{
  // freopen("in", "r", stdin);
    int sum, ans, needOne, needTwo;
    while(1)
    {
        sum = 0;
        for(int i = 1; i <= 6; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        if(!sum) break;
        ans = a[6] + a[5] + a[4] + (a[3] + 3) / 4;
        needOne = a[5] * 11 + one[a[3] % 4];
        needTwo = a[4] * 5 + two[a[3] % 4];
        if(needTwo <= a[2])
        {
            int left = a[2] - needTwo;
            int tmp = (left + 8) / 9;
            ans += tmp;
            needOne += 4 *(9 * tmp - left);
        }
        else
        {
            needOne += 4 * (needTwo - a[2]);
        }
        if(needOne <= a[1])
        {
            ans += (a[1] - needOne + 35) / 36;
        }
        printf("%d\n", ans);
    }
    return 0;
}

Fucker 2010-09-15 16:19 发表评论