• Educational Codeforces Round 42 (Rated for Div. 2)


    A. Equator(模拟)

    找权值的中位数,直接模拟。。

    代码写的好丑qwq。。

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    const int MAXN = 1e6 + 10;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N;
    int a[MAXN];
    main() { 
    #ifdef WIN32
    //    freopen("a.in", "r", stdin);
    #endif
        int N = read(), sum = 0;
        for(int i = 1; i <= N; i++) a[i] = read(), sum += a[i];
        int now = 0;
        for(int i = 1; i <= N; i++) {
            now += a[i];
            if(!(sum & 1)) {
                if(now >= sum / 2) {
                    printf("%d", i); return 0;
                }            
            } else {
                if(now > sum / 2) {
                    printf("%d", i); return 0;
                }    
            }
        }
    }
    A

    B. Students in Railway Carriage(贪心)

    直接贪心放,如果是偶数的话肯定是两种各放一半

    如果是奇数的话,应该在都放一半的基础上,把多的放在最后一个

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    const int MAXN = 1e6 + 10;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N;
    char s[MAXN];
    main() { 
    #ifdef WIN32
    //    freopen("a.in", "r", stdin);
    #endif
        int N = read(), A = read(), B = read();
        if(A < B) swap(A, B);
        scanf("%s", s + 1);
        int last = 0;
        int ans = 0; 
        s[N + 1] = '*';
        for(int i = 1; i <= N + 1; i++) {
            if(s[i] == '*') {
                int len = i - last - 1;
                ans += min(len / 2, A) + min(len / 2, B);
                A -= min(len / 2, A); B -= min(len / 2, B);
                if((len & 1) && A > 0) A--, ans++;
                if(A < B) swap(A, B);
                last = i;
            }
        }
        
        printf("%d", ans);
    }
    B

    C. Make a Square(数论,暴力)

    首先把所有平方数预处理出来

    然后对于输出的数的各个位分解出来,

    枚举所有的平方数,算出最少需要删几个

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #define int long long 
    using namespace std;
    const int MAXN = 1e6 + 10;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int po[MAXN], tot = 0, N, P[MAXN], Pnum = 0;
    int check(int val) {
        int times = 0;
        for(int i = 1; i <= Pnum; i++) {
            if(P[i] == val % 10) {
                val /= 10, times++;
            }
            if(val == 0) {
                return Pnum - times;
            }
        }
        if(val != 0) return -1;
        else return Pnum - times;
    }
    main() { 
    #ifdef WIN32
    //    freopen("a.in", "r", stdin);
    #endif
        for(int i = 1; ; i++) {
            po[++tot] = i * i;
            if(po[tot] > 2 * 1e9) break;
        }
        N = read();
        int x = N;
        while(x) P[++Pnum] = x % 10, x /= 10;
        int ans = 1e9;
        for(int i = 1; i <= tot; i++) {
            int num = check(po[i]);
            if(num != -1) ans = min(ans, num);
        }
        printf("%I64d", ans == 1e9 ? -1 : ans);
    }
    C

    D. Merge Equals(堆)

    用优先队列维护每个数的位置和权值(权值相同的时候位置小的在前面)

    然后每次取出前两个,如果相同就合成完再放回去,否则统计答案

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<queue>
    #define int long long 
    using namespace std;
    const int MAXN = 1e6 + 10;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N;
    struct Node {
        int pos, val;
        bool operator < (const Node &rhs) const {
            return val == rhs.val ? pos > rhs.pos : val > rhs.val;
        }
    };
    priority_queue<Node>q;
    int ans[MAXN];
    void work() {
        while(q.size() >= 2) {
            Node x = q.top(); q.pop();
            Node y = q.top(); q.pop();
            if(x.val != y.val) {q.push(y); ans[x.pos] = x.val; continue;}
            q.push((Node){y.pos, x.val * 2});
        }
        while(q.size() != 0) ans[q.top().pos] = q.top().val, q.pop();
    }
    main() { 
    #ifdef WIN32
    //    freopen("a.in", "r", stdin);
    #endif
        N = read();
        for(int i = 1; i <= N; i++)
            q.push((Node){i, read()});
        work();
        int num = 0;
        for(int i = 1; i <= N; i++)
            if(ans[i] != 0)
                num++;
        printf("%I64d
    ", num);
        for(int i = 1; i <= N; i++)
            if(ans[i] != 0)
                printf("%I64d ", ans[i]);
    }
    D
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9201470.html
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