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  • 河南省第五届ACM程序设计大赛

    D:   遥 控 器

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<queue>
    #include<vector>
    #include<map>
    using namespace std;
    typedef unsigned long long LL;
    #define met(a,b) (memset(a,b,sizeof(a)))
    const int INF = 1e9+7;
    const int N = 11000;
    const int MOD = 9973;
    
    int a[20], s, e;
    
    int Slove()
    {
        int i, Mintime = 100, shi=e/10, ge=e%10;
    
        if(s>e)
        {
            if(a[11])
                Mintime = min(Mintime, 100-(s-e));
            if(a[12])
                Mintime = min(Mintime, s-e);
        }
        else if(s<e)
        {
            if(a[11])
                Mintime = min(Mintime, e-s);
            if(a[12])
                Mintime = min(Mintime, 100-(e-s));
        }
        else
            Mintime = 0;
    
        if(s!=e && !a[11] && !a[12])
        {
            if(shi==0)
            {
                if(!a[ge]) Mintime = -1;
            }
            else
            {
                if(!a[10]) Mintime = -1;
                if(!a[shi]) Mintime = -1;
                if(!a[ge]) Mintime = -1;
            }
        }
    
        for(i=0; i<=99; i++)
        {
            if(i==s) continue;
    
            int sh = i/10, g = i%10;
            if(sh==0 && a[g])
            {
                if(i>e)
                {
                    if(a[11]) ///变大
                        Mintime = min(Mintime, 100-(i-e)+1);
                    if(a[12]) ///变小
                        Mintime = min(Mintime, (i-e)+1);
                }
                else if(i<e)
                {
                    if(a[11])
                        Mintime = min(Mintime, (e-i)+1);
                    if(a[12])
                        Mintime = min(Mintime, 100-(e-i)+1);
                }
                else Mintime = min(Mintime, 1);
            }
            else if(a[10] && a[sh] && a[g])
            {
                if(i>=e)
                {
                    if(a[11])
                        Mintime = min(Mintime, 100-(i-e)+3);
                    if(a[12])
                        Mintime = min(Mintime, (i-e)+3);
                }
                else if(i<e)
                {
                    if(a[11])
                        Mintime = min(Mintime, (e-i)+3);
                    if(a[12])
                        Mintime = min(Mintime, 100-(e-i)+3);
                }
                else Mintime = min(Mintime, 3);
            }
        }
    
        if(Mintime==100)
            Mintime = -1;
        return Mintime;
    }
    
    int main()
    {
        int T;
        scanf("%d", &T);
        while(T--)
        {
            met(a, 0);
    
            scanf("%d%d%d%d", &a[1], &a[2], &a[3], &a[11]);
            scanf("%d%d%d%d", &a[4], &a[5], &a[6], &a[12]);
            scanf("%d%d%d", &a[7], &a[8], &a[9]);
            scanf("%d%d", &a[10], &a[0]);
    
            scanf("%d%d", &s, &e);
    
            int ans = Slove();
    
            printf("%d
    ", ans);
        }
        return 0;
    }
    /**
    
    */
    View Code

    G:   Divideing Jewels (01背包)  代码写的有点丑

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<queue>
    #include<vector>
    #include<map>
    using namespace std;
    typedef unsigned long long LL;
    #define met(a,b) (memset(a,b,sizeof(a)))
    const int INF = 1e9+7;
    const int N = 11000;
    const int MOD = 9973;
     
    int a[20], dp[N], v[N], k;
     
    void Slove()  ///重点就在于怎样的把它转化为01背包
    {
        int i, j;
        k=1;
     
        for(i=1; i<=10; i++)
        for(j=1; j<=a[i]; j++)
               v[k++] = i;
        k--;
    }
     
    int main()
    {
     
        int iCase=1;
     
        while(scanf("%d%d%d%d%d%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5], &a[6], &a[7], &a[8], &a[9], &a[10])!=EOF)
        {
            int sum=0, i, j;
     
            for(i=1; i<=10; i++)
                sum += a[i]*i;
     
            if(sum==0) break;
     
            met(dp, 0);
            met(v, 0);
     
            Slove();
     
            for(i=1; i<=k; i++)
            for(j=sum/2; j>=v[i]; j--)
            {
                if(j>=v[i])
                    dp[j] = max(dp[j], dp[j-v[i]]+v[i]);
            }
     
     
            if(sum%2 || dp[sum/2]!=sum/2)
                printf("#%d:Can't be divided.
    
    ", iCase++);
            else
                printf("#%d:Can be divided.
    
    ", iCase++);
     
        }
     
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/YY56/p/5534950.html
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