• 2017 Multi-University Training Contest


    2017 Multi-University Training Contest - Team 5

    06 / hdu6090    思维

    tags: 主要是要想到,一个连通块中,最小化内部距离和,一定是一个点连剩下所有点。

    #include<bits/stdc++.h>
    using namespace std;
    #pragma comment(linker, "/STACK:102400000,102400000")
    #define rep(i,a,b) for (int i=a; i<=b; ++i)
    #define per(i,b,a) for (int i=b; i>=a; --i)
    #define mes(a,b)  memset(a,b,sizeof(a))
    #define INF 0x3f3f3f3f
    #define MP make_pair
    #define PB push_back
    #define fi  first
    #define se  second
    typedef long long ll;
    const int N = 200005;
    
    int main()
    {
        ll  n, m;
        int T;  scanf("%d", &T);
        while(T--)
        {
            scanf("%lld%lld", &n, &m);
            ll  ans=0;
            if(m<=n-1) {
                ans = m*m*2 + (n-(m+1))*(m+1)*n*2 + (n-(m+1))*(n-(m+2))*n;
            }
            else {
                ans = (n-1)*(n-1)*2 - (m-(n-1))*2;
                ans = max(ans, n*(n-1));
            }
            printf("%lld
    ", ans);
        }
    
        return 0;
    }

    08  / hdu6092    dp,类似反过来的背包

    tags: dp[now] 表示当前子集和为 now 的方案数,从1~m遍历过去,如果 dp[now] != b[now]  则需要更新,更新操作是:从 m~now ,dp[j] += dp[i-now] ,即类似背包反过来更新。

    #include<bits/stdc++.h>
    using namespace std;
    #pragma comment(linker, "/STACK:102400000,102400000")
    #define rep(i,a,b) for (int i=a; i<=b; ++i)
    #define per(i,b,a) for (int i=b; i>=a; --i)
    #define mes(a,b)  memset(a,b,sizeof(a))
    #define INF 0x3f3f3f3f
    #define MP make_pair
    #define PB push_back
    #define fi  first
    #define se  second
    typedef long long ll;
    const int N = 200005;
    
    int n, m, a[60], cnt;
    ll  b[N], dp[N];
    int main()
    {
        int T;  scanf("%d", &T);
        while(T--)
        {
            mes(dp, 0);  dp[0] = 1;
            cnt = 0;
            scanf("%d%d", &n, &m);
            rep(now,0,m)
            {
                scanf("%lld", &b[now]);
                while(dp[now]!=b[now])
                {
                    per(i,m,now) dp[i] += dp[i-now];
                    a[++cnt] = now;
                }
            }
            rep(i,1,n)
                if(i!=n) printf("%d ", a[i]);
                else printf("%d
    ", a[i]);
        }
    
        return 0;
    }

    11 / hdu6095   水

    #include<bits/stdc++.h>
    using namespace std;
    #pragma comment(linker, "/STACK:102400000,102400000")
    #define rep(i,a,b) for (int i=a; i<=b; ++i)
    #define per(i,b,a) for (int i=b; i>=a; --i)
    #define mes(a,b)  memset(a,b,sizeof(a))
    #define INF 0x3f3f3f3f
    #define MP make_pair
    #define PB push_back
    #define fi  first
    #define se  second
    typedef long long ll;
    const int N = 200005;
    
    int n, k, a[N];
    int main()
    {
        int T;  scanf("%d", &T);
        while(T--)
        {
            scanf("%d%d", &n, &k);
            rep(i,1,n) scanf("%d", &a[i]);
            sort(a+1, a+1+n);
            int cnt = 1;
            per(i,n-1,1)
            {
                if(a[i+1]-a[i]>k) break;
                else ++cnt;
            }
            printf("%d
    ", cnt);
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/sbfhy/p/7325807.html
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