河南省多校连萌（四）

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https://acm.zzuli.edu.cn/zzuliacm/contest.php?cid=1242

A
居然卡了map

#include<stdio.h>#include<math.h>#include<string.h>#include<vector>#include<queue>#include<map>#include<algorithm>using namespace std;const int INF  = 0x3f3f3f3f;const int MAXN = (int)5e5 + 5;int arr[MAXN];int main(){    int n;    while(scanf("%d", &n) != EOF) {    int cnt = 1;    for(int i = 0; i < n; ++i) {        scanf("%d", arr + i);    }    sort(arr, arr + n);    for(int i = 1; i < n; ++i) {        if(arr[i] == arr[i - 1]) continue;        ++cnt;    }    puts(cnt > n - cnt ? "Yes" : "No");    }    return 0;}

B
是个很有意思的题
最初的想法是反正是要去掉路径中最大的边，那么直接维护路径中最大边权，直接优先减去最大边权的路径距离跑Dijkstra
后来发现是不正确的，比如你从节点1出发到节点n去，现在在节点i，我们维护了路径 1 -> i 的最大边权，而有可能出现需要剪掉的最大边权在 i -> n 的路径中
所以1 和 n 为起点跑两遍最短路，在跑第二遍最短路的时候维护一下答案即可

#include<stdio.h>#include<string.h>#include<queue>#include<vector>#include<algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int MAXS = 60 * 1024 * 1024;char buf[MAXS], *ch;void read(int &x) {    while(*ch <= 32) ++ch;    for(x = 0; *ch >= '0'; ++ch) x = x * 10 + *ch - '0';}const int MAXN = (int)1e3 + 5;struct edge{int to, cost, nxt;} E[MAXN * MAXN]; int lnk[MAXN], sz;void add(int x, int y, int c) {    E[sz] = {y, c, lnk[x]};    lnk[x] = sz++;}int d[MAXN], dd[MAXN], MA[MAXN], n, m;struct node{int v, val, ma;};bool operator < (const node &x, const node &y) {    return x.val > y.val;}void slove() {    memset(d, 0x3f, sizeof(d));    memset(MA, 0, sizeof(MA));    d[1] = 0;    priority_queue<node> que;    que.push({1, 0, 0});    while(!que.empty()) {        node e = que.top(); que.pop();        if(d[e.v] < e.val) continue;        MA[e.v] = e.ma;        for(int i = lnk[e.v]; i; i = E[i].nxt) {            int dis = d[e.v] + E[i].cost + e.ma - max(e.ma, E[i].cost);            if(dis < d[E[i].to]) {                d[E[i].to] = dis;                que.push({E[i].to, dis, max(e.ma, E[i].cost)});            }        }    }    int ans = INF;    memset(dd, 0x3f, sizeof(dd));    dd[n] = 0;    que.push({n, 0, 0});    while(!que.empty()) {        node e = que.top(); que.pop();        if(dd[e.v] < e.val) continue;        ans = min(ans, e.val + e.ma + d[e.v]);        ans = min(ans, e.val + MA[e.v] + d[e.v]);        for(int i = lnk[e.v]; i; i = E[i].nxt) {            int dis = dd[e.v] + E[i].cost + e.ma - max(e.ma, E[i].cost);            if(dis < dd[E[i].to]) {                dd[E[i].to] = dis;                que.push({E[i].to, dis, max(e.ma, E[i].cost)});            }        }    }    if(ans == INF) puts("Impossible");    else printf("%d\n", ans);}int main(){    fread(buf, MAXS, 1, stdin);    ch = buf;    int T;    read(T);    while(T--) {        sz = 1;        memset(lnk, 0, sizeof(lnk));        read(n);        read(m);        for(int i = 0; i < m; ++i) {            int x, y, c;            read(x);            read(y);            read(c);            add(x, y, c);            add(y, x, c);        }    slove();    }    return 0;}

#include<stdio.h>#include<math.h>#include<string.h>#include<vector>#include<queue>#include<map>#include<algorithm>using namespace std;const int INF  = 0x3f3f3f3f;const int MAXN = (int)1e3 + 5;int arr[MAXN];int main(){    int T;    scanf("%d", &T);    while(T--) {    int n, m ;    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; ++i) {        scanf("%d", arr + i);    }    arr[++n] = 101;    int ans = 0;    for(int i = 0; i <= n; ++i) {        int low = arr[i];        int top = i + m + 1 <= n ? arr[i + m + 1] - 1 : 100;        ans = max(ans, top - low);    }     printf("%d\n", ans);    }    return 0;}

#include<stdio.h>#include<math.h>#include<string.h>#include<vector>#include<queue>#include<map>#include<algorithm>using namespace std;const int INF  = 0x3f3f3f3f;const int MAXN = (int)1e3 + 5;int main(){    int n;    while(scanf("%d", &n) != EOF) {        int cnt2 = 0, cnt4 = 0, cnt0 = 0;        for(int i = 0; i < n; ++i) {            int x;            scanf("%d", &x);            if(x % 4 == 0) ++cnt4;            else if(x % 2 == 0) ++cnt2;            else ++cnt0;        }        puts(cnt4 >= cnt0 || (cnt2 == 0 && cnt4 + 1 >= cnt0) ? "Pass" : "Not Pass");    }    return 0;}

E
这就是高中的等比数列求和，只是是在模意义下的，会模逆元就能AC了

#include<stdio.h>#include<math.h>#include<string.h>#include<vector>#include<queue>#include<map>#include<algorithm>using namespace std;const int INF  = 0x3f3f3f3f;const int MAXN = (int)1e3 + 5;const int MOD = (int)1e9 + 7;int modpow(int a, int n) {    int res = 1;    while(n) {    if(n&1) res = 1LL * res * a % MOD;        a = 1LL * a * a % MOD;        n >>= 1;    }    return res;}int main(){    int k, n, t = 0;    while(scanf("%d%d", &k, &n) != EOF) {        if(k == 1) {            printf("Case %d: %d\n", ++t, n + 1);            continue;        }        int ans = 1LL * (modpow(k, n + 1) - 1) % MOD * modpow(k - 1, MOD - 2) % MOD;        printf("Case %d: %d\n", ++t, ans);    }    return 0;}

F
n! 因子2 的个数

#include<stdio.h>#include<math.h>#include<string.h>#include<vector>#include<queue>#include<map>#include<algorithm>using namespace std;const int INF  = 0x3f3f3f3f;const int MAXN = (int)1e3 + 5;int main(){    int T;    scanf("%d", &T);    while(T--) {        int n;        scanf("%d", &n);        int ans = 0, t = 1;        while(n) {            n /= 2;            ans += n;        }        printf("%d\n", ans);    }    return 0;}

H
一个很重要的条件就是两个子序列长度为 n/2，可以利用这个条件加一个很强的剪枝，也就是说每个数值都要利用，那么用 vis[] 记录是否用过，第一个序列的下一个元素是之后没有用过的的第一个，然后再枚举第二个序列的下一个元素。

#pragma GCC optimize ("O2")#include<stdio.h>#include<string.h>int n, arr[50], vis[50], flag;void dfs(int p1, int p2, int cnt) {    if(flag || cnt == n) {        flag = 1;        return ;    }    int i = p1, j = p2;    for(; i <= n; ++i) if(!vis[i]) break;    if(i == n + 1) return ;    vis[i] = 1;    for(; j <= n; ++j) if(!vis[j] && arr[i] == arr[j]) {        vis[j] = 1;        dfs(i + 1, j + 1, cnt + 2);        vis[j] = 0;    }    vis[i] = 0;}int main(){    int T;    while(scanf("%d", &T) !=EOF  && T) {        while(T--) {            memset(vis, 0, sizeof(vis));            flag = 0;            scanf("%d", &n);            for(int i = 1; i <= n; ++i) {                scanf("%d", arr + i);            }            dfs(1, 1, 0);            printf(flag ? "竟然还有这种操作\n" : "没有这种操作\n");        }    }    return 0;}

I
二元函数 f(x,y)=ax+by，(x,y∈Z) 的值域为 {k⋅gcd(a,b)|k∈Z}
若 ax+by 要组成任何数，那么 gcd(a,b)≡1

下面是官方题解了：
每个都可以到达k的倍数，则
k*x-(2*n-2)*y=C (1<=C<=2*n-2)
必须有解，既gcd(k,2*n-2)==1

#include<stdio.h>int gcd(int a, int b) {    return b ? gcd(b, a % b) : a;}int main(){    int n, k;    while(scanf("%d%d", &n, &k) != EOF)        puts(gcd(2 * n - 2, k) != 1 ? "Yes" : "No");    return 0;}

后面的题会了再补上

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