Codeforces Round #250 (Div. 2)

题目链接

A：水题，暴力判断有没有3个两倍长或两倍短的字符串，注意如果有多个答案也是属于C的情况
B：先打出lowbit的表，然后从大到小去组合即可
C：贪心，权值大的点优先拿掉
D：并查集+贪心，先把边都按权值从大到小排序，然后加边，该边可以被用到的次数为左边集合个数乘上右边集合个数，最后答案在除以C2n即可

代码：
A：

#include <stdio.h>#include <string.h>char str[4][105];int main() {    int i, j, flag = 0;    char ans = 'C';    for (i = 0; i < 4; i++)        scanf("%s", str[i]);    for (i = 0; i < 4; i++) {        int s = 0, l = 0;        for (j = 0; j < 4; j++) {            if (i == j) continue;            int a = strlen(str[i]) - 2;            int b = strlen(str[j]) - 2;            if (a * 2 <= b)                s++;            if (a >= b * 2)                l++;                        }        if (s == 3 || l == 3) {            ans = i + 'A';            flag++;        }    }    if (flag != 1) ans = 'C';    printf("%c\n", ans);    return 0;}

B:

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 100001;int sum, limit, ans[N], an = 0;int i, j;struct Low {    int lowbit;    int num;} l[N];bool cmp(Low a, Low b) {    return a.lowbit > b.lowbit;}int main() {    scanf("%d%d", &sum, &limit);    for (i = 1; i < N; i++) {        l[i].num = i;        l[i].lowbit = (i&(-i));    }    sort(l + 1, l + N, cmp);    for (i = 1; i < N; i++) {        if (l[i].num <= limit) {            if (sum >= l[i].lowbit) {                ans[an++] = l[i].num;                sum -= l[i].lowbit;            }        }        if (sum == 0) break;    }    if (sum != 0) printf("-1\n");     else {        printf("%d\n", an);        for (int i = 0; i < an - 1; i++)            printf("%d ", ans[i]);        printf("%d\n", ans[an - 1]);    }    return 0;}

C:

#include <stdio.h>#include <string.h>#include <vector>#include <algorithm>#define max(a,b) ((a)>(b)?(a):(b))using namespace std;const int N = 1005;int n, m, vis[N], val[N];vector<int> g[N];struct Node {    int value;    int id;} node[N];bool cmp(Node a, Node b) {    return a.value > b.value;}int main() {    int i, j;    scanf("%d%d", &n, &m);    for (i = 1; i <= n; i++) {        scanf("%d", &node[i].value);        node[i].id = i;        val[i] = node[i].value;    }    int u, v;    while (m--) {        scanf("%d%d", &u, &v);        g[u].push_back(v);        g[v].push_back(u);    }    int ans = 0;    sort(node + 1, node + 1 + n, cmp);    for (i = 1; i <= n; i++) {        int u = node[i].id;        vis[u] = 1;        for (j = 0; j < g[u].size(); j++) {            int v = g[u][j];            if (vis[v]) continue;            ans += val[v];        }    }    printf("%d\n", ans);    return 0;}

D:

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 100005;int n, m, node[N], parent[N], sum[N];struct Edge {    int u, v, w;    Edge(int u = 0, int v = 0, int w = 0) {    this->u = u; this->v = v; this->w = w;    }} e[N];bool cmp(Edge a, Edge b) {    return a.w > b.w;}int find(int x) {    if (x == parent[x]) return x;    return parent[x] = find(parent[x]);}int main() {    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; i++) {    parent[i] = i;    sum[i] = 1;    scanf("%d", &node[i]);    }    int u, v;    for (int i = 0; i < m; i++) {    scanf("%d%d", &u, &v);    e[i] = Edge(u, v, min(node[u], node[v]));    }    sort(e, e + m, cmp);    double ans = 0;    for (int i = 0; i < m; i++) {    int pa = find(e[i].u);    int pb = find(e[i].v);    if (pa != pb) {        ans += (double)e[i].w * sum[pa] * sum[pb];        parent[pb] = pa;        sum[pa] += sum[pb];    }    }    printf("%.6lf\n", ans * 2 / (n * 1.0 * (n - 1)));    return 0;}