ACdream原创群赛(18)のAK‘s dream

题目链接

A：水题，直接模拟题意去搞即可

D：贪心+组合数学，把剑和英雄都从小到大排序，每次计算该英雄能用前几把剑cnt，cnt减去之前有多少人就是该英雄能选的，种数乘上cnt - 之前有多少人

F：DP，dp[i][j][2]表示第i天，用了j次修改，当前状态是A还是B，然后去转移即可

G：水题，直接用字符串处理即可，把数字都加上前导零方便判断

H：几何，利用正弦定理，余弦定理乱搞就可以了

I：水题，直接每个位置推出来输出，化简利用gcd即可

J：并查集，把询问逆向处理，每次多一块，就和左边和右边合并，这样对应的块数跟着变化，最后再反过来输出即可

代码：

A：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
 
int n, cur, d;
 
int main() {
    int cas = 0;
    while (~scanf("%d%d%d", &n, &cur, &d)) {
        printf("Case #%d: ", ++cas);
        int l = max(1, cur - d);
        if (cur == 1) printf("[<<]");
        else printf("(<<)");
        if (l != 1) printf("[...]");
        for (int i = l; i < cur; i++)
            printf("(%d)", i);
        printf("[%d]", cur);
        int r = min(n, cur + d);
        for (int i = cur + 1; i <= r; i++)
            printf("(%d)", i);
        if (r != n) printf("[...]");
        if (cur == n) printf("[>>]");
        else printf("(>>)");
        printf("\n");
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
 
const int N = 100005;
const int MOD = 1000000007;
typedef long long ll;
 
int t, n, a[N], b[N];
 
int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        for (int i = 0; i < n; i++)
            scanf("%d", &b[i]);
        sort(a, a + n);
        sort(b, b + n);
        int ans = 1;
        int l = 0;
        for (int i = 0; i < n; i++) {
            while (l < n && b[i] >= a[l]) l++;
            ans = (ll)ans * (l - i) % MOD;
        }
        printf("Case #%d: %d\n", ++cas, ans);
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
 
const int N = 105;
const int INF = 0x3f3f3f3f;
int n, m, a[N], Max[N][55][2], Min[N][55][2], num[2];
char str[N];
 
int main() {
    num[0] = 1;
    num[1] = -1;
    int cas = 0;
    while (~scanf("%d", &n)) {
        scanf("%s%d", str + 1, &m);
        memset(Max, -INF, sizeof(Max));
        memset(Min, INF, sizeof(Min));
        for (int i = 1; i <= n; i++)
            a[i] = str[i] - 'A';
        Max[0][0][0] = 0;
        Min[0][0][0] = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k < 2; k++) {
                    if (a[i + 1]) {
                        Max[i + 1][j][!k] = max(Max[i + 1][j][!k], Max[i][j][k]);
                        Max[i + 1][j + 1][k] = max(Max[i + 1][j + 1][k], Max[i][j][k] + num[k]);
                        Min[i + 1][j][!k] = min(Min[i + 1][j][!k], Min[i][j][k]);
                        Min[i + 1][j + 1][k] = min(Min[i + 1][j + 1][k], Min[i][j][k] + num[k]);
                    } else {
                        Max[i + 1][j + 1][!k] = max(Max[i + 1][j + 1][!k], Max[i][j][k]);
                        Max[i + 1][j][k] = max(Max[i + 1][j][k], Max[i][j][k] + num[k]);
                        Min[i + 1][j + 1][!k] = min(Min[i + 1][j + 1][!k], Min[i][j][k]);
                        Min[i + 1][j][k] = min(Min[i + 1][j][k], Min[i][j][k] + num[k]);
                    }
                }
            }
        }
        int ans1 = INF, ans2 = -INF;
        int st = m % 2;
        for (int i = st; i <= m; i += 2) {
            ans1 = min(ans1, Min[n][i][0]);
            ans1 = min(ans1, Min[n][i][1]);
            ans2 = max(ans2, Max[n][i][0]);
            ans2 = max(ans2, Max[n][i][1]);
        }
        int ans = -INF;
        ans = max(ans2, -ans1);
        printf("Case #%d: %d\n", ++cas, ans);
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
using namespace std;
 
string str;
 
int main() {
    while (cin >> str) {
        string a, b, c;
        a = "000000000000000000000000032767";
        b = "000000000000000000002147483647";
        c = "000000000009223372036854775807";
        int s = 0;
        int n = str.length();
        if (str[0] == '-') {
            n--;
            s++;
            a = "000000000000000000000000032768";
            b = "000000000000000000002147483648";
            c = "000000000009223372036854775808";
        }
        string tmp = "";
        for (int i = 30 - n; i > 0; i--)
            tmp += '0';
        for (int i = s; i < n + s; i++)
            tmp += str[i];
        str = tmp;
        if (str <= a) printf("short\n");
        else if (str <= b) printf("int\n");
        else if (str <= c) printf("long long\n");
        else printf("It is too big!\n");
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <cmath>
 
const double eps = 1e-9;
const double pi = acos(-1.0);
 
double a, b, c, d;
 
double solve() {
    if (fabs(a) < eps)
        return 0.0;
    if (fabs(c) < eps)
        return 0.0;
    if (fabs(b) < eps)
        return c;
    if (fabs(d) < eps)
        return b + c;
    a = a / 180 * pi; b = b / 180 * pi; c = c / 180 * pi; d = d / 180 * pi;
    double AB = 1.0;
    double AOB = pi - b - c;
    double ADB = pi - a - b - c;
    double AEB = pi - b - c - d;
    double A = a + b, B = c + d;
    double AO = AB * sin(c) / sin(AOB);
    double BO = AB * sin(b) / sin(AOB);
    double BD = AB * sin(A) / sin(ADB);
    double AE = AB * sin(B) / sin(AEB);
    double DO = BD - BO;
    double EO = AE - AO;
    double DE = sqrt(DO * DO + EO * EO - cos(AOB) * 2 * DO * EO);
    double cosx = (DE * DE + EO * EO - DO * DO) / 2 / DE / EO;
    double ans = acos(cosx) * 180 / pi;
    return ans;
}
 
int main() {
    while (~scanf("%lf%lf%lf%lf", &a, &b, &c, &d)) {
        printf("%.2lf\n", solve());
    }
    return 0;
}

#include <cstdio>
#include <cstring>
 
int gcd(int a, int b) {
    while (b) {
        int tmp = a % b;
        a = b;
        b = tmp;
    }
    return a;
}
 
int n;
 
int main() {
    while (~scanf("%d", &n)) {
        int a, b;
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &a, &b);
            int mu = b + 1;
            int d = gcd(a, mu);
            a /= d; mu /= d;
            if (mu < 0) {
                a = -a;
                mu = -mu;
            }
            if (mu == 1) printf("%d", a);
            else {
                printf("%d/%d", a, mu);
            }
            printf(" ");
            printf("%d", b + 1);
            printf("%c", i == n - 1 ? '\n' : ' ');
        }
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
 
const int N = 1000005;
 
int t, n, m, parent[N];
 
struct Block {
    int id, h;
} d[N];
 
int q[N], vis[N], out[N], on;
 
int find(int x) {
    return x == parent[x] ? x : parent[x] = find(parent[x]);
}
 
void merge(int u, int v) {
    int pa = find(u);
    int pb = find(v);
    if (pa != pb) parent[pb] = pb;
}
 
bool cmp(Block a, Block b) {
    return a.h > b.h;
}
 
int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        d[0].h = d[n + 1].h = -1;
        vis[0] = vis[n + 1] = 0;
        for (int i = 1; i <= n; i++) {
            vis[i] = 0;
            parent[i] = i;
            scanf("%d", &d[i].h);
            d[i].id = i;
        }
        sort(d + 1, d + n + 1, cmp);
        for (int i = 0; i < m; i++)
            scanf("%d", &q[i]);
        int l = 1;
        int ans = 0;
        on = 0;
        printf("Case #%d:", ++cas);
        for (int i = m - 1; i >= 0; i--) {
            while (l <= n && q[i] < d[l].h) {
                if (vis[d[l].id - 1] && vis[d[l].id + 1]) {
                    merge(d[l].id, d[l].id - 1);
                    merge(d[l].id, d[l].id + 1);
                    ans--;
                }
                else if (vis[d[l].id - 1])
                    merge(d[l].id, d[l].id - 1);
                else if (vis[d[l].id + 1])
                    merge(d[l].id, d[l].id + 1);
                else ans++;
                vis[d[l].id] = 1;
                l++;
            }
            out[on++] = ans;
        }
        for (int i = on - 1; i >= 0; i--)
            printf(" %d", out[i]);
        printf("\n");
    }
    return 0;
}

ACdream原创群赛(18)のAK's dream

原文：http://blog.csdn.net/accelerator_/article/details/39118333