BestCoder Round #71 (div.2)

 

数学 1001 KK's Steel

类似斐波那契求和

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>

typedef long long ll;
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;

ll run(ll n)    {
    ll a = 1, b = 2, c;
    ll sum = 3, ret = 2;
    while (sum < n) {
        c = a + b;
        if (sum + c < n)    {
            sum += c;
            a = b;  b = c;
            ret++;
        }
        if (sum + c > n)    return ret;
        if (sum + c == n)  return ret + 1;
    }
    return ret;
}

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        ll n;   scanf ("%I64d", &n);
        if (n <= 2) puts ("1");
        else    {
            ll ans = run (n);
            printf ("%I64d\n", ans);
        }
    }

    return 0;
}

数学 1002 KK's Point

圆上四个点连线能成圆内一个点，所以答案就是comb (n, 4) + n

#include <cstdio>

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        if (n < 4)  printf ("%d\n", n);
        else    {
            unsigned long long ans = 1;
            ans = ans * n * (n - 1) / 2 * (n - 2) / 3 * (n - 3) / 4 + n;
            printf ("%I64d\n", ans);
        }
    }

    return 0;
}

还有自己想出来的结论

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>

typedef long long ll;
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;

ll fun(int x)   {
    return 1ll * (1 + x) * x / 2;
}

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        ll ans = n;
        n -= 3;
        int t = 1;
        while (n >= 1)  {
            ans += fun (n) * t;
            n--;    t++;
        }
        printf ("%I64d\n", ans);
    }

    return 0;
}

DP 1004 KK's Number

显然，每个人的策略就是都会拿剩下的数中最大的某几个数

假如我们用f[i]表示当剩下i个数的时候先手得分-后手得分的最小值

那么得到f[i]=max\left(a[j+1]-f[j] \right)(1<j\leq i)f[i]=max(a[j+1]−f[j])(1<j≤i)

但是这样做，是要超时的

我们不妨简单转换一下 f[i]=_max; _max=max(_max,a[i+1]-f[i]);

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>

typedef long long ll;
const int N = 5e4 + 5;
const int MOD = 1e9 + 7;
int a[N];
long long dp[N];

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        for (int i=0; i<n; ++i) scanf ("%d", &a[i]);
        std::sort (a, a+n);
        dp[0] = a[0];
        for (int i=1; i<n; ++i) {
            dp[i] = std::max (dp[i-1], a[i] - dp[i-1]);
        }
        printf ("%I64d\n", dp[n-1]);
    }

    return 0;
}