2018年湖南省第十四届大学生计算机程序设计竞赛

目录

题目链接

传送门

题目

A题

思路

签到。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 200000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
int n;
 
int main() {
    scanf("%d", &n);
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo\n");
    printf("..o");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("o.o");
    for(int i = 1; i <= n; ++i) printf(".");
    printf(".o.");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("o.o\n");
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("o.o");
    for(int i = 1; i <= n; ++i) printf(".");
    printf(".o.");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo\n");
    printf("o..");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("o.o");
    for(int i = 1; i <= n; ++i) printf(".");
    printf(".o.");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("o.o\n");
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo");
    for(int i = 1; i <= n; ++i) printf(".");
    printf("ooo\n");
    return 0;
}

B题

思路

打表找规律。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 200000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
int n, m;
int C[4007][2007];
 
int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    for(int i = 0; i <= 4000; ++i) {
        C[i][0] = 1;
        for(int j = 1; j <= min(i, 2000); ++j) {
            C[i][j] = (C[i-1][j] + C[i-1][j-1]) % mod;
        }
    }
    while(~scanf("%d%d", &n, &m)) {
        printf("%lld\n", 1LL * (C[n+m][m] - 1) * (C[n+m][m] - 1) % mod);
    }
    return 0;
}

C题

思路

看样例猜。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 200000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
int h, c;
 
int main() {
    while(~scanf("%d%d", &h, &c)) {
        printf("%d\n", h > c ? h : c + 1);
    }
    return 0;
}

E题

思路

初始时联通快有\(n\times m\)个，由于在每次进行操作之后重新数联通块比较复杂，因此我们可以将思路转换一下，变成每次操作后联通块减少了\(x\)，那么答案就是\(n\times m-x\)。

我们可以发现每次增加一条新的横线(前面没有对这一行进行操作)后联通块减少了\(m-1\)个，每次增加一条竖线后联通块减少了\(n-1\)个。

我们假设横线数目为\(h\)，竖线数目为\(w\)，那么当只有横线时答案为\(n\times m-h\times(m-1)\)，当只有竖线时答案为\(n\times m-w\times(n-1)\)。当既有横线又有竖线时需要稍微容斥一下，因为对于交叉点我们会重复减掉，因此需要加回来，被重复减掉部分为\((h-1)\times(w-1)\)，此时答案为\(n\times m-h\times(m-1)-w\times(n-1)+(h-1)\times(w-1)\)。

注意同一行被操作多次只能被算一次，因此我们可以考虑用左闭右开线段树来维护。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 200000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
bool lazy[3][maxn<<2];
int n, m, q, cnt;
int op[maxn], l[maxn], r[maxn], sum[3][maxn<<2], num[maxn*2];
 
void push_down(int rt, int l, int r, int id) {
    if(!lazy[id][rt]) return;
    lazy[id][rt] = 0;
    lazy[id][lson] = lazy[id][rson] = 1;
    int mid = (l + r) >> 1;
    sum[id][lson] = num[mid+1] - num[l];
    sum[id][rson] = num[r+1] - num[mid+1];
}
 
void push_up(int rt, int id) {
    sum[id][rt] = sum[id][lson] + sum[id][rson];
}
 
void build(int rt, int l, int r) {
    for(int i = 0; i < 2; ++i) sum[i][rt] = 0, lazy[i][rt] = false;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(lson, l, mid), build(rson, mid + 1, r);
}
 
void update(int rt, int l, int r, int L, int R, int id) {
    if(l == L && r == R) {
        sum[id][rt] = num[R+1] - num[L];
        lazy[id][rt] = true;
        return;
    }
    push_down(rt, l, r, id);
    int mid = (l + r) >> 1;
    if(R <= mid) update(lson, l, mid, L, R, id);
    else if(L > mid) update(rson, mid + 1, r, L, R, id);
    else {
        update(lson, l, mid, L, mid, id);
        update(rson, mid + 1, r, mid + 1, R, id);
    }
    push_up(rt, id);
}
 
int getnum(int x) {
    return lower_bound(num + 1, num + cnt + 1, x) - num;
}
 
int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    while(~scanf("%d%d%d", &n, &m, &q)) {
        cnt = 0;
        for(int i = 1; i <= q; ++i) {
            scanf("%d%d%d", &op[i], &l[i], &r[i]);
            num[++cnt] = l[i], num[++cnt] = r[i] + 1;
        }
        sort(num + 1, num + cnt + 1);
        cnt = unique(num + 1, num + cnt + 1) - num - 1;
        build(1, 1, cnt);
        for(int i = 1; i <= q; ++i) {
            l[i] = getnum(l[i]), r[i] = getnum(r[i]+1);
            if(op[i] == 1) {
                update(1, 1, cnt, l[i], r[i]-1, 0);
            } else {
                update(1, 1, cnt, l[i], r[i]-1, 1);
            }
            // printf("[%d %d]\n", sum[0][1], sum[1][1]);
            LL ans = 1LL * n * m - 1LL * sum[0][1] * (m - 1) - 1LL * sum[1][1] * (n - 1);
            if(sum[0][1] && sum[1][1]) ans += 1LL * (sum[0][1] - 1) * (sum[1][1] - 1);
            printf("%lld\n", ans);
        }
    }
    return 0;
}

H题

思路

由于查询时保证\(r-l\leq 2\)，因此我们可以用三个树状数组分别存下以\(i\)为左端点，区间长度为\(0,1,2\)的区间被哪些区间覆盖过。

插入时对于\([l,r]\)的每个数作为左端点区间长度为\(0\)的均有被覆盖，\([l,r-1]\)为左端点区间长度为\(1\)的均有被覆盖，\([l,r-2]\)为左端点区间长度为\(2\)的均有被覆盖，分别进行\(update\)即可。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 100000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
inline int read() {
    int f = 0, x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
 
int n, q, op, l, r;
int tree[5][maxn];
 
void bit_add(int x, int id, int val) {
    while(x <= n) {
        tree[id][x] += val;
        x += lowbit(x);
    }
}
 
int bit_query(int x, int id) {
    int ans = 0;
    while(x) {
        ans += tree[id][x];
        x -= lowbit(x);
    }
    return ans;
}
 
int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    while(~scanf("%d%d", &n, &q)) {
        for(int i = 0; i <= 2; ++i) {
            for(int j = 0; j <= n; ++j) tree[i][j] = 0;
        }
        while(q--) {
            op = read(), l = read(), r = read();
            if(op == 1) {
                bit_add(l, 0, 1);
                bit_add(r + 1, 0, -1);
                if(l <= r - 1) {
                    bit_add(l, 1, 1);
                    bit_add(r, 1, -1);
                }
                if(l <= r - 2) {
                    bit_add(l, 2, 1);
                    bit_add(r - 1, 2, -1);
                }
            } else {
                printf("%d\n", bit_query(l, r-l));
            }
        }
    }
    return 0;
}

J题

思路

我们考虑当前\(dfs\)到结点\(u\)，从\(1\)到\(u\)的父亲结点得到的答案为\(sum\)，那么\(f_u=sum+u\)的贡献。

当\(a[u]\)在\(1\)到\(u\)这条链上是第一次出现时，此时\(u\)的贡献为这条链上不同数的个数；

当\(a[u]\)是第二次出现时，此时\(u\)的贡献为上一次\(a[u]\)出现位置到\(u\)的路径上不同数的个数\(+1\),这个\(1\)表示的有序对是\((a[u],a[u])\)。

当\(a[u]\)出现次数大于\(2\)时，此时\(u\)的贡献为上一次\(a[u]\)出现位置到\(u\)的路径上不同数的个数。

在回溯的时候记得清除\(u\)造成的影响。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 200000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
int ans[maxn], sum;
int n, u, tot, cnt;
int head[maxn], a[maxn], vis[maxn], lst[maxn];
 
struct edge {
    int v, next;
}ed[maxn*2];
 
void add(int u, int v) {
    ed[tot].v = v;
    ed[tot].next = head[u];
    head[u] = tot++;
}
 
void dfs(int u) {
    int tmp = 0;
    if(!vis[a[u]]) tmp = cnt, ++cnt;
    else if(vis[a[u]] == 1) tmp = cnt - lst[a[u]] + 1;
    else tmp = cnt - lst[a[u]];
    ans[u] = sum + tmp;
    int pp = lst[a[u]];
    lst[a[u]] = cnt;
    sum += tmp;
    ++vis[a[u]];
    for(int i = head[u]; ~i; i = ed[i].next) {
        int v = ed[i].v;
        dfs(v);
    }
    sum -= tmp;
    --vis[a[u]];
    if(vis[a[u]] == 0) --cnt;
    lst[a[u]] = pp;
}
 
int main() {
    while(~scanf("%d", &n)) {
        tot = cnt = sum = 0;
        assert(n >= 1 && n <= 100000);
        for(int i = 1; i <= n; ++i) vis[i] = ans[i] = lst[i] = 0, head[i] = -1;
        for(int i = 2; i <= n; ++i) {
            scanf("%d", &u);
            assert(u >= 1 && u <= n && u != i);
            add(u, i);
        }
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            assert(a[i] >= 1 && a[i] <= n);
        }
        dfs(1);
        for(int i = 2; i <= n; ++i) printf("%d\n", ans[i]);
    }
    return 0;
}

K题

思路

我们将\(a\)做个前缀和，然后考虑每个\(b\)的贡献即可。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define fi first
#define se second
#define lson (rt<<1)
#define rson (rt<<1|1)
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 500000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n, m, L, R;
LL sum[maxn], a[maxn], b[maxn];
 
int main() {
    while (~scanf("%d%d%d%d", &n, &m, &L, &R)) {
        L += 2, R += 2;
        sum[0] = 0;
        for (int i = 1; i <= n + 1; i++) scanf("%lld", &a[i]), sum[i] = sum[i - 1] + a[i];
        for (int i = 1; i <= m + 1; i++) scanf("%lld", &b[i]);
        LL ans = 0;
        for (int i = 1; i <= m + 1; i++) {
            int l = max(1, L - i), r = min(n + 1, R - i);
            if (l > n + 1 || r < 1) continue;
            ans = (ans + 1LL * b[i] * ((sum[r] - sum[l - 1] + mod) % mod) % mod) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}
posted @ 2019-10-04 10:20  Dillonh  阅读(760)  评论(0)  编辑  收藏  举报