A
统计 A 会做但 B 不会做的题目数量和 B 会做但 A 不会做的数量，无解就是 A 会做但 B 不会做的题目数量为 (0)，输出 -1，有解就输出这两个的和除以 A 会做但 B 不会做的题目数。
#include <bits/stdc++.h>
#define reg register
#define ll long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define SZ(x) (int) x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
using namespace std;
const int MAXN = 110;
int n, r[MAXN], b[MAXN];
inline void work() {
scanf("%d", &n);
int count1 = 0, count2 = 0;
for(reg int i = 1; i <= n; ++i)
scanf("%d", &r[i]);
for(reg int i = 1; i <= n; ++i)
scanf("%d", &b[i]);
for(reg int i = 1; i <= n; ++i)
if(r[i] && !b[i])
++count1;
for(reg int i = 1; i <= n; ++i)
if(!r[i] && b[i])
++count2;
if(count1 == 0) {
puts("-1");
return;
}
printf("%d
", (count1 + count2) / count1);
}
int main() {
// freopen("input.txt", "r", stdin);
work();
return 0;
}
B
(b_j - b_i = j - i) 就等于 (b_j - j = b_i - i)。
然后用 map 维护一下就行了。
#include <bits/stdc++.h>
#define reg register
#define ll long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define SZ(x) (int) x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
using namespace std;
const int MAXN = 2e5 + 10;
int n, b[MAXN];
map<int, ll> mp;
inline void work() {
scanf("%d", &n);
for(reg int i = 0; i < n; ++i) {
scanf("%d", &b[i]);
mp[b[i] - i] += b[i];
}
ll ans = 0;
for(auto it : mp)
ans = max(ans, it.second);
printf("%lld
", ans);
}
int main() {
// freopen("input.txt", "r", stdin);
work();
return 0;
}
C
统计每个字符出现次数，枚举删除的字符，统计答案。
#include <bits/stdc++.h>
#define reg register
#define ll long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define er erase
#define SZ(x) (int) x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
using namespace std;
char ch, B[1 << 20], *S = B, *T = B;
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T) ? 0 : *S++)
#define isd(c) (c >= '0' && c <= '9')
int aa, bb;
int rd() {
while(ch = getc(), !isd(ch) && ch != '-');
ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
while(ch = getc(), isd(ch))
aa = aa * 10 + ch - '0';
return bb ? aa : -aa;
}
const int MAXN = 210;
int n, vis[MAXN];
string s;
inline void work() {
cin >> n;
memset(vis, 0, sizeof(vis));
for(reg int i = 1; i <= n; ++i) {
char ch;
cin >> ch;
s += ch;
vis[ch]++;
}
int ans = 0;
for(reg char ch = 'z'; ch >= 'b'; ch--) {
if(!vis[ch])
continue;
for(reg int i = 0; i < SZ(s); ++i)
if(s[i] == ch) {
if((i + 1 < SZ(s) && s[i] - 1 == s[i + 1]) || (i - 1 >= 0 && s[i] - 1 == s[i - 1])) {
s.er(s.begin() + i);
i--;
ans++;
}
}
for(reg int i = SZ(s) - 1; i >= 0; i--) {
if(s[i] == ch) {
if((i + 1 < SZ(s) && s[i] - 1 == s[i + 1]) || (i - 1 >= 0 && s[i] - 1 == s[i - 1])) {
s.er(s.begin() + i);
i++;
ans++;
}
}
}
}
printf("%d
", ans);
}
int main() {
// freopen("input.txt", "r", stdin);
work();
return 0;
}
D
预处理每一个点到达终点的最短距离，然后分类讨论，求最少次数，就是如果下一个位置可以是导航导出来的 ans 就不变，否则 ans+1。求最多次数，就是下一个位置只要不是唯一的最短路，ans 就加一。
#include <bits/stdc++.h>
#define reg register
#define ll long long
#define ull unsigned long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define er erase
#define SZ(x) (int) x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
using namespace std;
char ch, B[1 << 20], *S = B, *T = B;
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T) ? 0 : *S++)
#define isd(c) (c >= '0' && c <= '9')
int aa, bb;
int rd() {
while(ch = getc(), !isd(ch) && ch != '-');
ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
while(ch = getc(), isd(ch))
aa = aa * 10 + ch - '0';
return bb ? aa : -aa;
}
const int MAXN = 2e5 + 10;
vi F[MAXN], G[MAXN];
int n, m, k, p[MAXN], s, t, dis[MAXN], ans1, ans2;
queue<int> q;
inline void bfs() {
q.push(p[k]);
dis[p[k]] = 1;
while(!q.empty()) {
int x = q.front();
q.pop();
for(reg int i = 0; i < SZ(G[x]); ++i) {
int v = G[x][i];
if(!dis[v]) {
dis[v] = dis[x] + 1;
q.push(v);
}
}
}
}
inline void solve_min(int u) {
if(u == k)
return;
int x = p[u];
if(dis[p[u + 1]] == dis[x] - 1)
solve_min(u + 1);
else {
++ans1;
solve_min(u + 1);
}
}
inline void solve_max(int u) {
if(u == k)
return;
int x = p[u];
if(dis[p[u + 1]] != dis[x] - 1) {
++ans2;
solve_max(u + 1);
return;
}
for(reg int i = 0; i < SZ(F[x]); ++i) {
int v = F[x][i];
if(v != p[u + 1] && dis[v] == dis[x] - 1) {
++ans2;
solve_max(u + 1);
return;
}
}
solve_max(u + 1);
}
inline void work() {
n = rd();
m = rd();
for(reg int i = 1; i <= m; ++i) {
int x = rd(), y = rd();
F[x].pb(y);
G[y].pb(x);
}
k = rd();
for(reg int i = 1; i <= k; ++i)
p[i] = rd();
bfs();
solve_min(1);
solve_max(1);
printf("%d %d
", ans1, ans2);
}
int main() {
// freopen("input.txt", "r", stdin);
work();
return 0;
}
E
二维偏序问题，将一维从小到大排序，另一维线段树维护最大值。
#include <bits/stdc++.h>
#define reg register
#define ll long long
#define int long long
#define ull unsigned long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define er erase
#define SZ(x) (int) x.size()
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
using namespace std;
char ch, B[1 << 20], *S = B, *T = B;
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T) ? 0 : *S++)
#define isd(c) (c >= '0' && c <= '9')
int aa, bb;
int rd() {
while(ch = getc(), !isd(ch) && ch != '-');
ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
while(ch = getc(), isd(ch))
aa = aa * 10 + ch - '0';
return bb ? aa : -aa;
}
const int MAXN = 2e5 + 10;
int n, m, p, mx[MAXN << 2], tag[MAXN << 2], d[MAXN];
struct Node {
int x, y;
} a[MAXN], b[MAXN];
struct Equ {
int x, y, z;
} c[MAXN];
bool cmp1(Node a, Node b) {
return a.x < b.x;
}
bool cmp2(Equ a, Equ b) {
return a.x < b.x;
}
inline void pushup(int x) {
mx[x] = max(mx[x << 1], mx[x << 1 | 1]);
}
inline void pushdown(int x) {
if(tag[x]) {
tag[x << 1] += tag[x];
tag[x << 1 | 1] += tag[x];
mx[x << 1] += tag[x];
mx[x << 1 | 1] += tag[x];
tag[x] = 0;
}
}
inline void build(int l, int r, int x) {
if(l == r) {
mx[x] = -b[l].y;
return;
}
int mid = (l + r) >> 1;
build(l, mid, x << 1);
build(mid + 1, r, x << 1 | 1);
pushup(x);
}
inline void modify(int l, int r, int ql, int qr, int x, int c) {
if(ql <= l && qr >= r) {
tag[x] += c;
mx[x] += c;
return;
}
pushdown(x);
int mid = (l + r) >> 1;
if(ql <= mid)
modify(l, mid, ql, qr, x << 1, c);
if(qr > mid)
modify(mid + 1, r, ql, qr, x << 1 | 1, c);
pushup(x);
}
int query(int l, int r, int ql, int qr, int x) {
if(ql <= l && qr >= r)
return mx[x];
pushdown(x);
int mid = (l + r) >> 1;
int ret = -numeric_limits<int>::max();
if(ql <= mid)
ret = max(ret, query(l, mid, ql, qr, x << 1));
if(qr > mid)
ret = max(ret, query(mid + 1, r, ql, qr, x << 1 | 1));
return ret;
}
inline void work() {
n = rd();
m = rd();
p = rd();
for(reg int i = 1; i <= n; ++i) {
a[i].x = rd();
a[i].y = rd();
}
for(reg int i = 1; i <= m; ++i) {
b[i].x = rd();
b[i].y = rd();
}
for(reg int i = 1; i <= p; ++i) {
c[i].x = rd();
c[i].y = rd();
c[i].z = rd();
}
sort(a + 1, a + n + 1, cmp1);
sort(b + 1, b + m + 1, cmp1);
sort(c + 1, c + p + 1, cmp2);
for(reg int i = 1; i <= m; ++i)
d[i] = b[i].x;
build(1, m, 1);
int pt = 1, ans = -numeric_limits<int>::max();
for(reg int i = 1; i <= n; ++i) {
while(pt <= p && c[pt].x < a[i].x) {
int pos = ub(d + 1, d + m + 1, c[pt].y) - d;
if(pos <= m)
modify(1, m, pos, m, 1, c[pt].z);
pt++;
}
ans = max(ans, query(1, m, 1, m, 1) - a[i].y);
}
printf("%lld
", ans);
}
signed main() {
// freopen("input.txt", "r", stdin);
work();
return 0;
}
F
首先我们可以发现每次变换相当于把 (0) 向左或向右移动两位，移动前后相邻两个 (0) 之间 (1) 的个数奇偶不变，那么我们将这两个串中的 (0) 都尽量往某方向移动，如果移动之后两串长得一样，则这两原串间可相互变换。所以我们将每个 (0) 的权值设为与前一个 (0) 之间 (1) 个数的奇偶（也就是设成 digit % 2，其中 digit 表示前面所说的 (1) 的个数，然后用哈希判断是否相同即可。
#include <bits/stdc++.h>
#define reg register
#define ll long long
#define ull unsigned long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define er erase
#define SZ(x) (int) x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
using namespace std;
char ch, B[1 << 20], *S = B, *T = B;
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T) ? 0 : *S++)
#define isd(c) (c >= '0' && c <= '9')
int aa, bb;
int rd() {
while(ch = getc(), !isd(ch) && ch != '-');
ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
while(ch = getc(), isd(ch))
aa = aa * 10 + ch - '0';
return bb ? aa : -aa;
}
const int MAXN = 2e5 + 10;
const int BASE = 233;
char s[MAXN];
int n, a[MAXN], tot, flag[MAXN], cnt[MAXN];
ull power[MAXN], Hash[MAXN];
ull get(int l, int r) {
int len = r - l + 1;
return 1ll * Hash[r] - Hash[l - 1] * power[len];
}
ull check(int l, int r) {
int L = lower_bound(a + 1, a + tot + 1, l) - a;
int R = upper_bound(a + 1, a + tot + 1, r) - a - 1;
if(L <= R) {
ull x = (a[L] - l) & 1, v = x * power[R - L];
if(L < R)
v += get(L + 1, R);
return v;
}
return 0;
}
inline void work() {
power[0] = 1;
for(reg int i = 1; i < MAXN; ++i)
power[i] = power[i - 1] * BASE;
scanf("%d%s", &n, s + 1);
int digit = 0;
for(reg int i = 1; i <= n; ++i) {
cnt[i] = cnt[i - 1];
if(s[i] == '1')
digit++;
else {
flag[++tot] = digit & 1;
digit = 0;
a[tot] = i;
++cnt[i];
}
}
for(reg int i = 1; i <= tot; ++i)
Hash[i] = Hash[i - 1] * BASE + flag[i];
int q;
scanf("%d", &q);
while(q--) {
int l1, l2, len, r1, r2;
scanf("%d%d%d", &l1, &l2, &len);
r1 = l1 + len - 1;
r2 = l2 + len - 1;
if(cnt[r1] - cnt[l1 - 1] == cnt[r2] - cnt[l2 - 1] && check(l1, r1) == check(l2, r2))
printf("Yes
");
else
printf("No
");
}
}
int main() {
// freopen("input.txt", "r", stdin);
work();
return 0;
}