来自学长的推荐
A. Set
还做过类似的题，但是真的忘了
发现该题特殊点在于有 \(n\) 个数，求能被 \(n\) 整除的，明明是两个看似无关的数据，却给了同一个值，那么这里就是解题的关键
我们维护前缀和，最多有 \(n\)种取值， 如果前缀和为 \(0\) 那么从一开始到这个位置就是一个合法解，所以我们有一个默认的 \(0\) 的前缀和， 这样我们有 \(n + 1\) 个前缀和，而任意两个前缀和相同就是一组解，根据鸽巢原理，一定有解
code
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1000005;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9')c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return x;
}
int n, a[maxn], s[maxn];
int las[maxn];
int main(){
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
int n = read();
for(int i = 1; i <= n; ++i)a[i] = read() % n;
for(int i = 1; i <= n; ++i)s[i] = (s[i - 1] + a[i]) % n;
int pos = 0;
for(int i = 1; i <= n; ++i)if(s[i] == 0){pos = i;break;}
if(pos){
printf("%d\n",pos);
for(int i = 1; i <= pos; ++i)printf("%d ",i);
return 0;
}
for(int i = 1; i <= n; ++i)if(las[s[i]]){pos = i; break;}else las[s[i]] = i;
int l = las[s[pos]] + 1, r = pos;
printf("%d\n",r - l + 1);
for(int i = l; i <= r; ++i)printf("%d ",i);
return 0;
}
B. Read
读错题，，，，，，垃圾题暴零， 我就是sb
给的书是无序的。。。。
那么就是找是否存在大于等于\((n + 1) / 2\)的书，直接摩尔投票
code
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxm = 1005;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9')c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return x;
}
int m, k, cnt[maxm], x[maxm], y[maxm], z[maxm];
signed main(){
freopen("b.in", "r", stdin);
freopen("b.out", "w", stdout);
m = read(); k = read();
int sn = 0;
for(int i = 1; i <= m; ++i)cnt[i] = read();
for(int i = 1; i <= m; ++i)sn += cnt[i];
for(int i = 1; i <= m; ++i)x[i] = read();
for(int i = 1; i <= m; ++i)y[i] = read();
for(int i = 1; i <= m; ++i)z[i] = read();
int s = (1 << k) - 1;
int now, las = -1, ct = 0;
for (int i = 1; i <= m; ++i) {
now = x[i];
if(ct && las != now)--ct;
else las = now, ++ct;
long long opt = x[i];
for (int j = 1; j < cnt[i]; ++j) {
opt = (opt * y[i] + z[i]) & s;
now = opt;
if(ct && las != now)--ct;
else las = now, ++ct;
}
}
int sum = 0;
for (int i = 1; i <= m; ++i) {
now = x[i];
if(now == las)++sum;
long long opt = x[i];
for (int j = 1; j < cnt[i]; ++j) {
opt = (opt * y[i] + z[i]) & s;
now = opt;
if(now == las)++sum;
}
}
printf("%d\n",max(0, sum - sn + sum - 1));
return 0;
}
C. 题目交流通道
如果不考虑边权为 \(0\) 的，那么找出不经过 \(i - > j\) 的最短路径，判断与 \(i - > j\) 路径的大小关系简单维护答案
考场写出来，也发现会有算重的情况，但是并没有发现是因为边权为 \(0\)
现在考虑有边权为 \(0\) 的， 他们之间其实可以看做一个团， 答案对团内团外分别处理
对于团外，应该说是两个团之间， 我们用类似上面部分分的做法找出不经过 \(i - > j\) 的最短路径，注意这里是两个团，不是两个点，我们只能路过其他团，维护答案分情况讨论
如果该路径等于两团直接路径，那么这些路径可以在限制以上随便选 \((k - dis_{i, j} + 1) ^ {size_i \times size_j}\)
如果大于， 那么两团之间的路径只要不都是大于两团直接路径的即可， \((k - dis_{i, j} + 1) ^ {size_i \times size_j} - (k - dis_{i, j}) ^ {size_i \times size_j}\)
对于团内的我们用如下公式计算
设\(g_n\)表示 \(n\) 个点的图的方案数， \(f_n\) 表示 \(n\) 个点的团的方案数，
显然有\(g_n = (k + 1) ^ {(^n_2)}\)
那么\(f_n = g_n \sum_{i = 1} ^ {n - 1} f_i \times g_{n - i} \times (^{n - 1}_{i - 1}) \times k ^{i (n - i)}\)跟游乐园那个同理
code
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9')c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return x;
}
typedef long long ll;
const int maxn = 405;
const int inf = 2147483647;
const int mod = 998244353;
int n, k;
int mp[maxn][maxn], f[maxn], g[maxn],c[maxn][maxn];
bool vis[maxn];
ll d[maxn][maxn];
int qpow(int x, int y){
int ans = 1;
for(; y; y >>= 1, x = 1ll * x * x % mod)if(y & 1)ans = 1ll * ans * x % mod;
return ans;
}
struct SET{
int size[maxn],f[maxn];
void pre(){
for(int i = 1; i <= n; ++i)size[i] = 1;
for(int i = 1; i <= n; ++i)f[i] = i;
}
int fa(int x){return f[x] = f[x] == x ? x : fa(f[x]);}
void hb(int x, int y){
x = fa(x), y = fa(y);
if(x != y){
if(size[x] < size[y]){
size[y] += size[x];
f[x] = y;
}else{
size[x] += size[y];
f[y] = x;
}
}
}
}s;
int work(){
for(int i = 1; i <= n; ++i)
for(int j = i + 1; j <= n; ++j)
if(mp[i][j] != mp[j][i] || mp[i][j] > k)return 0;
for(int i = 1; i <= n; ++i)if(mp[i][i])return 0;
s.pre();
for(int i = 1; i <= n; ++i)
for(int j = i + 1; j <= n; ++j)
if(mp[i][j] == 0)s.hb(i, j);
for(int i = 1; i <= n; ++i)g[i] = qpow(k + 1, i * (i - 1) / 2);
for(int i = 0; i <= n; ++i){
c[i][0] = 1;
for(int j = 1; j <= i; ++j)
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
}
f[1] = 1;
for(int i = 2; i <= n; ++i){
f[i] = g[i];
for(int j = 1; j <= i - 1; ++j){
f[i] = (f[i] - 1ll * f[j] * g[i - j] % mod * c[i - 1][j - 1] % mod * qpow(k, j * (i - j)) % mod + mod) % mod;
}
}
memset(d,0x3f,sizeof(d));
for(int k = 1; k <= n; ++k){
if(s.f[k] != k)continue;
for(int i = 1; i <= n; ++i){
if(s.f[i] != i || i == k)continue;
for(int j = 1; j <= n; ++j){
if(s.f[j] != j || i == j || k == j)continue;
d[i][j] = min(d[i][j], (ll)mp[i][k] + mp[k][j]);
}
}
}
int ans = 1;
for(int i = 1; i <= n; ++i){
if(s.f[i] != i)continue;
for(int j = i + 1; j <= n; ++j){
if(s.f[j] != j || s.f[j] == s.f[i])continue;
if(d[i][j] < mp[i][j])return 0;
if(d[i][j] > mp[i][j])ans = 1ll * ans * (0ll + qpow(k - mp[i][j] + 1, s.size[i] * s.size[j]) - qpow(k - mp[i][j], s.size[i] * s.size[j])% mod) % mod;
else ans = 1ll * ans * qpow(k - mp[i][j] + 1, s.size[i] * s.size[j]) % mod;
}
}
for(int i = 1; i <= n; ++i)if(s.f[i] == i)ans = 1ll * ans * f[s.size[i]] % mod;
return ans;
}
int main(){
freopen("c.in", "r", stdin);
freopen("c.out", "w",stdout);
n = read(), k = read();
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
mp[i][j] = read();
printf("%d\n",work());
return 0;
}
D. 题目难度提升
从中位数往前找是否有相同的，有的话扔两个，再一个最大的一个较小的扔即可
没有的话，先把最小的扔进去，然后用对顶堆维护中位数
考虑加数的过程，要求中位数一直小于等于手中的最小值，
然后通过已经加入数的数量的奇偶判断一下是否需要放最小值，或者找最大的能放的
所谓对顶堆就是大根堆维护较小值，小根堆维护最大值
code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
using namespace std;
const int maxn = 100005;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9')c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return x;
}
int a[maxn], ans[maxn];
int n;
bool check(){
sort(a + 1,a + n + 1);
int pos = (n + 1) / 2;
while(pos > 1){
if(a[pos] == a[pos - 1]){
while(a[pos + 1] == a[pos]) ++pos;
ans[1] = ans[2] = a[pos];
int pl = pos - 2, pr = n, p = 2;
while(pl > 0 && pr > pos){
ans[++p] = a[pr]; --pr;
ans[++p] = a[pl]; --pl;
}
while(pr > pos)ans[++p] = a[pr--];
while(pl > 0)ans[++p] = a[pl--];
return true;
}
--pos;
}
return false;
}
multiset<int>s;
priority_queue<int> qmax;
priority_queue<int, vector<int> , greater<int> > qmin;
void push(int x){
if(qmax.empty() || x < qmax.top())qmax.push(x);
else qmin.push(x);
if(qmax.size() < qmin.size())qmax.push(qmin.top()), qmin.pop();
if(qmax.size() > qmin.size() + 1)qmin.push(qmax.top()), qmax.pop();
}
void work(){
ans[1] = a[1];
for(int i = 2; i <= n; ++i)s.insert(a[i]);
push(a[1]);
int p = 1;
while(!s.empty()){
int mi = *s.begin(), now = 0;
if(qmax.size() == qmin.size()){
if(mi >= qmin.top())now = *--s.end();
else now = mi;
}else{
if(!qmin.empty() && mi + mi >= qmax.top() + qmin.top())now = *--s.end();
else now = *--s.upper_bound(mi + mi - qmax.top());
}
ans[++p] = now;
push(now);
s.erase(now);
}
}
int main(){
freopen("d.in","r",stdin);
freopen("d.out","w",stdout);
n = read();
for(int i = 1; i <= n; ++i)a[i] = read();
sort(a + 1, a + n + 1);
if(check() == false)work();
for(int i = 1; i <= n; ++i)printf("%d ",ans[i]);printf("\n");
return 0;
}
