Codeforces Round #342 (Div. 2)

贪心 A - Guest From the Past

先买塑料和先买玻璃两者取最大值

#include <bits/stdc++.h>

typedef long long ll;

int main(void)  {
    ll n, a, b, c; std::cin >> n >> a >> b >> c;
    ll ans = 0;
    if (n >= a) {
        ll cnt1 = n / a;
        ll res = n % a;
        if (res >= b)   {
            cnt1 += (res - b) / (b - c) + 1;
        }
        ans = std::max (ans, cnt1);
    }
    if (n >= b) {
        ll cnt2 = (n - b) / (b - c) + 1;
        cnt2 += (n - cnt2 * (b - c)) / a;
        ans = std::max (ans, cnt2);
    }
    std::cout << ans << '\n';

    return 0;
}

暴力 B - War of the Corporations

#include <bits/stdc++.h>

const int N = 1e5 + 5;
char str1[N], str2[33];

int main(void)  {
    scanf ("%s%s", &str1, &str2);
    int len1 = strlen (str1);
    int len2 = strlen (str2);
    int ans = 0;
    for (int i=0; i<len1; ++i)  {
        if (str1[i] == str2[0]) {
            bool flag = true;
            for (int ii=i+1, j=1; j<len2; ++ii, ++j)  {
                if (str1[ii] != str2[j])    {
                    flag = false;   break;
                }
            }
            if (flag)   {
                ans++;  i += len2 - 1;
            }
        }
    }
    printf ("%d\n", ans);

    return 0;
}

构造 C - K-special Tables

#include <bits/stdc++.h>

int a[502][502];

int main(void)  {
    int n, k;   std::cin >> n >> k;
    int now = n * n;
    int mx = 0;
    for (int i=1; i<=n; ++i)    {
        for (int j=n; j>=k; --j)    {
            a[i][j] = now--;
            if (j == k) mx += a[i][j];
        }
    }
    for (int i=1; i<=n; ++i)    {
        for (int j=k-1; j>=1; --j)  {
            a[i][j] = now--;
        }
    }
    std::cout << mx << '\n';
    for (int i=1; i<=n; ++i)    {
        for (int j=1; j<=n; ++j)    {
            std::cout << a[i][j] << ' ';
        }
        std::cout << '\n';
    }

    return 0;
}

构造 D - Finals in arithmetic

题意：一个数字a + 反过来的ar == n (<=10^100000)，已知n，求a

分析：完全看别人代码看懂的。不考虑进位的话，那么n应该是是回文的。那么处理成不进位的，每一位0~18。

#include <bits/stdc++.h>

const int N = 1e5 + 5;
char str[N];
char ans[N];
int s[N];
int n;

bool judge(void)    {
    for (int i=0; i<n/2;)   {
        int l = i, r = n - 1 - i;
        if (s[l] == s[r])   ++i;
        else if (s[l] == s[r] + 1 || s[l] == s[r] + 11) {
            s[l]--; s[l+1] += 10;
        }
        else if (s[l] == s[r] + 10) {
            s[r] += 10; s[r-1]--;
        }
        else    return false;
    }
    if (n % 2 == 1)  {
        if (s[n/2] % 2 == 1 || s[n/2] > 18 || s[n/2] < 0)   return false;
        ans[n/2] = s[n/2] / 2 + '0';
    }
    for (int i=0; i<n/2; ++i)   {
        if (s[i] > 18 || s[i] < 0)  return false;
        ans[i] = (s[i] + 1) / 2 + '0';
        ans[n-1-i] = s[i] / 2 + '0';
    }
    return ans[0] > '0';
}

int main(void)  {
    scanf ("%s", str);
    n = strlen (str);
    for (int i=0; i<n; ++i) s[i] = str[i] - '0';
    if (judge ())   printf ("%s\n", ans);
    else if (str[0] == '1' && n > 1) {
        for (int i=0; i<n; ++i) {
            s[i] = str[i+1] - '0';
        }
        s[0] += 10; n--;
        if (judge ())   printf ("%s\n", ans);
        else    puts ("0");
    }
    else    puts ("0");

    return 0;
}