Codeforces Round #348(VK Cup 2016 - Round 2)

 

A - Little Artem and Presents (div2)

1 2 1 2这样加就可以了

#include <bits/stdc++.h>

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

int main() {
    int n; scanf ("%d", &n);
    int ans = n / 3 * 2;
    if (n % 3) {
        ans++;
    }
    printf ("%d\n", ans);
    return 0;
}

B - Little Artem and Grasshopper (div2)

水题，暴力模拟一下

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
char str[N];
int a[N];

int main() {
    int n; scanf ("%d", &n);
    scanf ("%s", str);
    for (int i=0; i<n; ++i) {
        scanf ("%d", a+i);
    }
    int now = 0;
    while (true) {
        if (now < 0 || now >= n) {
            break;
        }
        if (a[now] == -1) {
            puts ("INFINITE");
            return 0;
        }
        if (str[now] == '>') {
            int pre = now;
            now = now + a[now];
            a[pre] = -1;
        } else {
            int pre = now;
            now = now - a[now];
            a[pre] = -1;
        }
    }
    puts ("FINITE");
    return 0;
}

构造 C - Little Artem and Matrix (div2)

倒过来做，循环也反着来

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e2 + 5;
const int Q = 1e4 + 5;
int a[N][N];
int t[Q], row[Q], col[Q], x[Q];

int main() {
    int n, m, q; scanf ("%d%d%d", &n, &m, &q);
    for (int i=1; i<=q; ++i) {
        scanf ("%d", t+i);
        if (t[i] == 1) {
            scanf ("%d", row+i);
        }
        if (t[i] == 2) {
            scanf ("%d", col+i);
        }
        if (t[i] == 3) {
            scanf ("%d%d%d", row+i, col+i, x+i);
        }
        //printf ("%d %d %d %d\n", t[i], row[i], col[i], x[i]);
    }
    for (int i=q; i>=1; --i) {
        if (t[i] == 1) {
            int last = a[row[i]][m];
            for (int j=m; j>=2; --j) {
                a[row[i]][j] = a[row[i]][j-1];
            }
            a[row[i]][1] = last;
        }
        if (t[i] == 2) {
            int last = a[n][col[i]];
            for (int j=n; j>=2; --j) {
                a[j][col[i]] = a[j-1][col[i]];
            }
            a[1][col[i]] = last;
        }
        if (t[i] == 3) {
            a[row[i]][col[i]] = x[i];
        }
    }
    for (int i=1; i<=n; ++i) {
        for (int j=1; j<=m; ++j) {
            printf ("%d%c", a[i][j], j == m ? '\n' : ' ');
        }
    }
    return 0;
}

数学 D - Little Artem and Dance (div2)

题意：男生与女生围成圈跳舞，女生的位置不变，男生可以移动x个女生或者相邻的男生奇偶互换，问最后男生的排列

分析：问题的关键点在于奇数男生的圈顺序不变，偶数也不变，只是起点的位置改变，所以只要对两个起点操作就行了。

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e6 + 5;
int ans[N];

int main() {
    int p0 = 0, p1 = 1;
    int n, q; scanf ("%d%d", &n, &q);
    for (int i=0; i<q; ++i) {
        int type; scanf ("%d", &type);
        if (type == 1) {
            int x; scanf ("%d", &x);
            p0 = (p0 + x + n) % n;
            p1 = (p1 + x + n) % n;
        } else {
            p0 = p0 ^ 1;
            p1 = p1 ^ 1;
        }
    }
    for (int i=0; i<n; i+=2) {
        ans[(p0+i)%n] = i + 1;
    }
    for (int i=1; i<n; i+=2) {
        ans[(p1+i-1)%n] = i + 1;
    }
    for (int i=0; i<n; ++i) {
        printf ("%d%c", ans[i], i == n-1 ? '\n' : ' ');
    }
    return 0;
}

数学+前（后）缀 C - Little Artem and Random Variable (div1)

题意：已知p(max(a,b)=k) 和 p(min(a,b)=k)的概率，求p(a=k) 和 p(b=k)

分析：

P(a = k) = P(a <= k) — P(a <= k-1) P(max(a, b) <= k) = P(a <= k) * P(b <= k)

P(min(a, b) >= k) = P(a >= k) * P(b >= k) = (1 — P(a <= k-1)) *(1 — P(b <= k-1))

即

　

解方程的和，从而求得和

#include <bits/stdc++.h>

const int N = 1e5 + 5;
double p[N], q[N], a[N], b[N];

int main() {
    int n; scanf ("%d", &n);
    for (int i=1; i<=n; ++i) {
        scanf ("%lf", p+i);
        p[i] += p[i-1];
    }
    for (int i=1; i<=n; ++i) {
        scanf ("%lf", q+i);
    }
    for (int i=n; i>=1; --i) {
        q[i] += q[i+1];
    }
    for (int i=1; i<=n; ++i) {
        double A = p[i], B = q[i+1];
        double C = B - A - 1;
        double delta = sqrt (std::max (C*C - 4 * A, 0.0));
        a[i] = (-C+delta) / 2;
        b[i] = (-C-delta) / 2;
    }
    for (int i=1; i<=n; ++i) {
        printf ("%.10f%c", a[i] - a[i-1], i == n ? '\n' : ' ');
    }
    for (int i=1; i<=n; ++i) {
        printf ("%.10f%c", b[i] - b[i-1], i == n ? '\n' : ' ');
    }
    return 0;
}

　　

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