Codechef October Challenge 2019 Div.2
CodeChef = ccf
S10E
水题，直接模拟即可
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
int a[100010];
int main() {
int t = read();
while(t--) {
int n = read(), ans = 0;
for(int i = 1; i <= n; ++i) {
a[i] = read();
int minn = 2147483647;
for(int j = max(i-5, 1); j <= i-1; ++j)
minn = min(minn, a[j]);
if(minn > a[i]) ++ans;
}
printf("%d\n", ans);
}
return 0;
}
SAKTAN
行和列分开处理，因为奇数+偶数=奇数，所以分别求出加了偶数次的行\(ea\)和列\(eb\)，加了奇数次的行\(oa\)和列\(ob\)最后答案就是\(oa\times eb + ea \times ob\)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
int a[100010], b[100010];
int main() {
int t = read();
while(t--) {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
int n = read(), m = read(), q = read();
for(int i = 1; i <= q; ++i) {
int x = read(), y = read();
++a[x], ++b[y];
}
LL oa = 0, ob = 0, ea = 0, eb = 0;
for(int i = 1; i <= n; ++i)
if(a[i] & 1) ++oa;
else ++ea;
for(int i = 1; i <= m; ++i)
if(b[i] & 1) ++ob;
else ++eb;
LL ans = ob * ea + oa * eb;
printf("%lld\n", ans);
}
return 0;
}
MARM
每\(3n\)次操作是一个循环。但如果\(n\)是奇数，需特判最中间的数，当\(k \geq (n+1)/2\)时，最中间的数恒为\(0\)
不要忘记开long long！
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
LL a[10010];
int main() {
int t = read();
while(t--) {
LL n = read(), k = read();
for(int i = 0; i < n; ++i) a[i] = read();
LL flag = 0;
if(k >= n/2ll+1ll) flag = 1;
k = k % (3ll * n);
for(LL i = 0; i <= k-1; ++i)
a[i%n] ^= a[n-(i%n)-1ll];
for(LL i = 0; i < n; ++i) {
if((n & 1ll) && (i == n/2ll) && flag)
printf("0 ");
else printf("%lld ", a[i]);
}
printf("\n");
}
return 0;
}
MSV
对每个数暴力分解因数，然后用桶记录每个因数出现了几次
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
int tong[1000010], ans = 0;
int main() {
int t = read();
while(t--) {
int n = read(); ans = 0;
memset(tong, 0, sizeof(tong));
for(int i = 1; i <= n; ++i) {
int x = read();
ans = max(tong[x], ans);
++tong[1]; ++tong[x];
for(int j = 2; j * j <= x; ++j) {
if(x % j == 0) {
++tong[j];
if(x/j != j) ++tong[x/j];
}
}
}
printf("%d\n", ans);
}
return 0;
}
EVEDG
- 如果图的边数为偶数，则将所有点都划分到一个集合里即可
- 如果为边数为奇数
- 如果图中有一个点的度数为奇数，将这个点单独划为一个集合，其他点为另一个集合
- 如果所有点的度数都为偶数，先将一个有出度的点划为一个集合，然后就会变成上面的情况，即有至少一个点的度数会变成奇数。此情况需要三个集合
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
int du[100010];
struct zzz {
int t, nex;
}e[100010 << 1]; int head[100010], tot;
inline void add(int x, int y) {
e[++tot].t = y;
e[tot].nex = head[x];
head[x] = tot;
}
int main() {
int t = read();
while(t--) {
int n = read(), m = read(); tot = 0;
for(int i = 1; i <= n; ++i) du[i] = 0, head[i] = 0;
for(int i = 1; i <= m; ++i) {
int x = read(), y = read();
++du[x], ++du[y];
add(x, y); add(y, x);
}
if(!(m & 1)) {
printf("1\n");
for(int i = 1; i <= n; ++i)
printf("1 ");
}
else {
bool flag = 0;
for(int i = 1; i <= n; ++i)
if(du[i] & 1) flag = 1;
if(flag) {
flag = 0;
printf("2\n");
for(int i = 1; i <= n; ++i)
if((du[i] & 1) && flag == 0) printf("2 "), flag = 1;
else printf("1 ");
}
else {
int pos = 0;
for(int x = 1; x <= n; ++x)
if(du[x]) {
for(int i = head[x]; i; i = e[i].nex) --du[e[i].t];
pos = x; break;
}
printf("3\n");
for(int i = 1; i <= n; ++i)
if(i == pos) printf("3 ");
else if((du[i] & 1) && flag == 0) printf("2 "), flag = 1;
else printf("1 ");
}
}
printf("\n");
}
return 0;
}
MSNG
直接暴力模拟就好了
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define ULL unsigned long long
#define lim 1000000000000ll
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
struct zzz {
int len, maxx; LL opt, a[51];
}x[110];
int n;
bool judge(LL k) {
if(k > lim) return 0;
for(int i = 2; i <= n; ++i) {
bool flag = 0;
for(LL j = x[i].maxx; j <= 36; ++j) {
LL a = 0;
for(int o = 1; o <= x[i].len; ++o) {
a = a * j + x[i].a[o];
if(a > lim) return 0;
}
if(a == k) {
flag = 1; break;
}
}
if(!flag) return 0;
}
return 1;
}
int main() {
int t = read();
while(t--) {
n = read();
memset(x, 0, sizeof(x));
for(int i = 1; i <= n; ++i) {
x[i].opt = read();
char s[51]; scanf("%s", s+1);
int len = strlen(s+1);
x[i].len = len;
for(int j = 1; j <= len; ++j) {
if(s[j] >= '0' && s[j] <= '9')
x[i].a[j] = s[j] - '0';
else x[i].a[j] = s[j] - 'A' + 10;
x[i].maxx = max((LL)x[i].maxx, x[i].a[j] + 1);
}
}
for(int i = 2; i <= n; ++i) {
if(x[1].opt != -1) break;
if(x[i].opt != -1) {
swap(x[1], x[i]);
break;
}
if(x[i].maxx > x[1].maxx) swap(x[1], x[i]);
else if(x[i].maxx == x[1].maxx && x[i].len > x[1].len) swap(x[1], x[i]);
else if(x[i].maxx == x[1].maxx && x[i].len == x[1].len) {
for(int j = 1; j <= x[i].len; ++j)
if(x[i].a[j] > x[1].a[j]) {
swap(x[1], x[i]); break;
}
}
}
LL k = 0;
if(x[1].opt != -1) {
for(int i = 1; i <= x[1].len; ++i) {
k = k * x[1].opt + x[1].a[i];
if(k > lim) break;
}
if(!judge(k)) printf("-1\n");
else printf("%lld\n", k);
}
else {
bool flag = 0;
for(LL o = x[1].maxx; o <= 36; ++o) {
k = 0;
for(int i = 1; i <= x[1].len; ++i) {
k = k * o + x[1].a[i];
if(k > lim) break;
}
//cout << k << endl;
if(judge(k)) {
printf("%lld\n", k); flag = 1;
break;
}
}
if(!flag) printf("-1\n");
}
}
return 0;
}
BACREP
每过一秒，点权会下移。设点\(x\)的深度为\(deth_x\)，一个修改操作的时间是\(t\)，被修改的点\(i\)的深度为点\(deth_i\)，当前时间是\(T\)，只有当\(deth_x - T = deth_i - t\)、且\(x\)为\(i\)的后代时时，点\(x\)才会受这次修改影响。叶子节点需要特殊处理，它会受到所有\(deth_x - T \leq deth_i - t\)的修改的影响
考虑如何用线段树来维护\(deth_x - T\)
将操作离线，然后进行\(dfs\)，对每个节点\(x\)，枚举和它有关的所有操作，设该操作进行的时间为\(t\)。如果为修改，就在\(deth_x - t\)的位置加上修改的值\(k\)。如果是非叶子结点的询问，直接查询\(deth_x - t\)。如果是叶子节点的询问，则查询区间\([deth_x-t,MAXN]\)
在回溯时要将所有的修改操作撤回，防止对其他子树造成影响
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define LL long long
#define ls p << 1
#define rs p << 1 | 1
#define mid ((l+r) >> 1)
#define eps 500005
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
char read_c() {
char c = getchar();
while(c != '?' && c != '+') c = getchar();
return c;
}
struct zzz {
int t, nex;
}e[500010 << 1]; int head[500010], tot;
struct hhh {
bool opt; LL k, tim;
};
vector <hhh> que[500010];
inline void add(int x, int y) {
e[++tot].t = y;
e[tot].nex = head[x];
head[x] = tot;
}
bool leaf[500010];
int f[500010], deth[500010];
LL v[500010], tree[1000020 << 2];
void dfs(int x, int fa) {
f[x] = fa, deth[x] = deth[fa] + 1; leaf[x] = 1;
for(int i = head[x]; i; i = e[i].nex) {
if(e[i].t != fa) dfs(e[i].t, x);
else continue;
leaf[x] = 0;
}
}
inline void up(int p) {
tree[p] = tree[ls] + tree[rs];
}
LL query(int pos, int l = 1, int r = 1000010, int p = 1) {
if(l == r) return tree[p];
if(pos <= mid) return query(pos, l, mid, ls);
else return query(pos, mid+1, r, rs);
}
LL sum(int nl, int nr, int l = 1, int r = 1000010, int p = 1) {
if(l >= nl && r <= nr) return tree[p];
LL anss = 0;
if(nl <= mid) anss += sum(nl, nr, l, mid, ls);
if(nr > mid) anss += sum(nl, nr, mid+1, r, rs);
return anss;
}
void update(int pos, LL k, int l = 1, int r = 1000010, int p = 1) {
if(l == r) {
tree[p] += k; return ;
}
if(pos <= mid) update(pos, k, l, mid, ls);
else update(pos, k, mid+1, r, rs);
up(p);
}
LL ans[500010]; bool vis[500010];
void dfs2(int x, int fa) {
int len = que[x].size();
update(deth[x] + eps, v[x]);
for(int i = 0; i < len; ++i) {
hhh a = que[x][i];
if(a.opt == 1) {
int pos = deth[x] - a.tim + eps;
update(pos, a.k);
}
else {
vis[a.tim] = 1;
if(!leaf[x])
ans[a.tim] = query(deth[x] - a.tim + eps);
else {
ans[a.tim] = sum(deth[x] - a.tim + eps, 1000010);
}
}
}
for(int i = head[x]; i; i = e[i].nex)
if(e[i].t != fa) dfs2(e[i].t, x);
update(deth[x] + eps, -v[x]);
for(int i = 0; i < len; ++i) {
hhh a = que[x][i];
if(a.opt == 1) {
int pos = deth[x] - a.tim + eps;
update(pos, -a.k);
}
}
}
int main() {
int n = read(), q = read();
for(int i = 1; i <= n-1; ++i) {
int x = read(), y = read();
add(x, y); add(y, x);
}
for(int i = 1; i <= n; ++i) v[i] = read();
dfs(1, 0);
for(int i = 1; i <= q; ++i) {
char opt = read_c();
int x = read();
if(opt == '+') que[x].push_back((hhh){1, read(), i});
else que[x].push_back((hhh){0, -1, i});
}
dfs2(1, 0);
for(int i = 1; i <= q; ++i)
if(vis[i]) printf("%lld\n", ans[i]);
return 0;
}
MAXLIS
真不会，乱搞只搞到了\(4.4\)分……
