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  • 暑假第十二测

    题解:第一题:

    打表找规律,从12以后每次加49,我一直在前十找规律,,找了半天,以后还是多打点,学聪明点

    #include <bits/stdc++.h>
    #define ll long long
    using namespace std;
    int zl[4] = {1,5,10,50};
    ll ans[] = {0,4,10,20,35,56,83,116,155,198,244,292,341,390,439,488,537,586,635,684,733,782,831,880,929,978,1027,1076,1125,1174,1223,1272,1321,1370,1419,1468,1517,1566,1615,1664,1713,1762,1811,1860,1909,1958,2007,2056,2105,2154,2203,2252,2301,2350,2399,2448,2497,2546,2595,2644,2693,2742,2791,2840,2889,2938,2987,3036,3085,3134,3183,3232,3281,3330,3379,3428,3477,3526,3575,3624,3673,3722,3771,3820,3869,3918,3967,4016,4065,4114,4163,4212,4261,4310,4359,4408,4457,4506,4555,4604,4653};
    int Ans = 0;
    int main()
    {    
        freopen("a.in","r",stdin);
        freopen("a.out","w",stdout);
        ll n;
        scanf("%I64d", &n);
        if(n <= 12)printf("%I64d
    ", ans[n]);
        else printf("%I64d
    ", 49* (n - 12) + ans[12]);
    }
    View Code

    第二题:

    这是一道傻逼题,但我就是没想出来,利用log单增函数就好了,比较log2 x^y. log2 y^x '即 ylog2x  xlog2y

    #include <bits/stdc++.h>
    using namespace std;
    
    int main()
    {
        freopen("b.in","r",stdin);
        freopen("b.out","w",stdout);
        int T;
        scanf("%d", &T);
        while(T--){
            double a, b;
            scanf("%lf%lf", &a, &b);
               if(a == b){puts("=");continue;}
               if(a == 1 && b > 1){puts("<");continue;}
            if(b == 1 && a > 1){puts(">");continue;}
            if(a*log(b) > b*log(a)){puts("<");continue;}
            if(a*log(b) < b*log(a)){puts(">");continue;}
            puts("=");
            //printf("%d
    ", work(a, b, x, y, aa, bb));
       }
        return 0;
    }
    View Code

    第三题:

    这道题最开始看着像上司的舞会,但简单画画图就知道会成环,然后就二分图匹配,原题是bzoj数字匹配,把数按质因子个数奇偶拆成两组,最大点独立集

    #include <bits/stdc++.h>
    #define ll long long
    using namespace std;
    const int M = 500005;
    int a[M], tmp[M];
    bool check(int a, int b){
        if(a > b)swap(a, b);
        if(b%a)return 0;
        int d = b/a;
        for(int i = 2; i <= d; i++){
            if(d%i == 0 && d != i)return 0;
        }
        return 1;    
    }
    struct {int id, val;}odd[M], even[M];
    int h[M], match[M], tot, cnt, t2, t1;
    bool vis[M];
    struct edge{int v,nxt;}G[M<<1];
    void add(int u, int v){G[++tot].nxt=h[u]; h[u]=tot; G[tot].v=v;}
    void init(){
        memset(h, 0, sizeof(h));
        memset(match, 0, sizeof(match));
        tot = cnt = t1= t2 = 0;
        
    }
    
    bool find(int u){
        for(int i = h[u]; i; i = G[i].nxt){
            int v = G[i].v;
            if(vis[v])continue;
            vis[v] = 1;
            if(!match[v] || find(match[v])){
                match[v] = u;
                return 1;
            }
        }
        return 0;
    }
    
    int main()
    {
        freopen("c.in","r",stdin);
        freopen("c.out","w",stdout);
        int T;
        scanf("%d", &T);
        while(T--){
            int n, ans = 0, cur = 0;
            scanf("%d", &n);
            init();
            for(int i = 1; i <= n; i++){
                scanf("%d", &a[i]);
                int tt = a[i];
                int cnt = 0;
                for(int j = 2; j <= a[i]; j++)
                    if(a[i]%j == 0){
                        while(a[i]%j == 0) a[i] /= j, cnt++;
                    }
                if(cnt & 1) odd[++t1].val = tt, odd[t1].id = ++cur;
                else even[++t2].val = tt, even[t2].id = ++cur;
            }
            for(int i = 1; i <= t1; i++)
                for(int j = 1; j <= t2; j++)
                    if(check(odd[i].val, even[j].val))add(odd[i].id, even[j].id);
            for(int i = 1; i <= t1; i++){
                memset(vis, 0, sizeof(vis));
                if(find(odd[i].id)) ans++;
            }
                
            printf("%d
    ", n - ans);
            
       }
        return 0;
    }
    View Code
    早上洛谷说忌模拟赛,结果真的倒数,而且今天题挺水的,自己太菜了
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  • 原文地址:https://www.cnblogs.com/EdSheeran/p/9460615.html
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