题目链接
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("*********
")
#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
");
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
");
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
");
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
");
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
");
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("*********
")
#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
", 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("*********
")
#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
", h > c ? h : c + 1);
}
return 0;
}
E题
思路
初始时联通快有(n imes 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 imes 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 imes m-h imes(m-1))，当只有竖线时答案为(n imes m-w imes(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) imes(w-1))，此时答案为(n imes m-h imes(m-1)-w imes(n-1)+(h-1) imes(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("*********
")
#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]
", 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
", ans);
}
}
return 0;
}
H题
思路
由于查询时保证(r-lleq 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("*********
")
#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
", 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("*********
")
#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
", 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("*********
")
#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
", ans);
}
return 0;
}