题解是我“氪金”弄来的，勿外传。。否则会被官方封号的
customize football jersey
nfl shop coupon code
nfl steelers
custom nfl jersey
nike air jordan blue
wig store
adidas ultraboost
nfl jersey sales
adidas yeezy boost 350 v2 bone
customized jerseys
nfl steelers
human hair wig
nike air jordan
custom football jerseys
sex toys
注意：Div1的同学是CDEF四道题目，Div2的同学是ABCD四道题目。
A题：可持久化动态仙人掌的直径问题（雾）
题解思路：
本题代码（简单到爆，，，所以叫签到题）
#include<bits/stdc++.h>
int n,m;
int main(){
scanf("%d%d",&n,&m);
printf("%d",(int)(pow(n,1.0/m)+1e-12));
return 0;
}Continue to next page....
B题：混凝土数学
题解思路：
本题代码：
#include<bits/stdc++.h>
using namespace std;
#define FL(a) {freopen(a".in","r",stdin);freopen(a".out","w",stdout);}
#define Mod 998244353
#define int long long
const int mx=200000;
int n,a,tax[200005],sum[200005];
const int inv2=(Mod+1)/2,inv3=(Mod+1)/3,inv6=inv2*inv3%Mod;
signed main(){
// FL("data1-1");
scanf("%lld",&n);
for(int i=1;i<=n;++i){
scanf("%lld",&a);
tax[a]++;
}
int ans=0;
for(int i=1;i<=mx;++i)sum[i]=(sum[i-1]+tax[i])%Mod;
for(int i=1;i<=mx;++i){
int N=tax[i];
if(N<2)continue;
// cerr<<"N: "<<N<<endl;
ans=(ans+N*(N-1)/2%Mod*(sum[min(i*2-1,mx)]-N)%Mod+N*(N-1)%Mod*(N-2)%Mod*inv6%Mod)%Mod;
}
printf("%lld\n",ans);
cerr<<"ans: "<<ans<<endl;
return 0;
}
C题：论如何玩转 Excel 表格
思路：
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<map>
using namespace std;
const int maxn = 2e6+10;
int read(){
int x=0,f=1;
char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=(x<<3)+(x<<1)+c-48;c=getchar();}
return x*f;
}
int k,m,n;
int a[maxn],b[maxn],c[maxn],d[maxn],fa[maxn],p[maxn];
pair<int,int> P[maxn];
int sum[maxn];
void add(int x){for(int i=x;i<=n;i+=i&-i)sum[i]++;}
int query(int x){int SUM=0;for(int i=x;i;i-=i&-i)SUM+=sum[i];return SUM;}
int main(){
n=read();
for(int i=1;i<=n;i++)a[i]=read();
for(int i=1;i<=n;i++)b[i]=read();
for(int i=1;i<=n;i++){c[i]=read();P[c[i]]=make_pair(i,1);}
for(int i=1;i<=n;i++){d[i]=read();P[d[i]]=make_pair(i,2);}
for(int i=1;i<=n;i++){fa[a[i]]=b[i];fa[b[i]]=a[i];}
for(int i=1;i<=n;i++){
if(fa[c[i]]!=d[i] || fa[d[i]]!=c[i]){puts("dldsgay!!1");return 0;}
if((P[a[i]].first+P[a[i]].second)%2!=(i+1)%2){puts("dldsgay!!1");return 0;}
if((P[b[i]].first+P[b[i]].second)%2!=(i+2)%2){puts("dldsgay!!1");return 0;}
}
for(int i=1;i<=n;i++)p[a[i]]=i;
for(int i=1;i<=n;i++)c[i]=p[c[i]]+p[d[i]];
long long ans=0;
for(int i=n;i>=1;i--){
ans+=query(c[i]);
add(c[i]);
}
printf("%lld\n",ans);
return 0;
}D题：可重集
思路：
代码：
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define INT __int128
#define ULL unsigned long long
int read(){
int a=0;char c=getchar();
while(c>'9'||c<'0')c=getchar();
while('0'<=c&&c<='9'){
a=a*10+c-48;
c=getchar();
}
return a;
}
int MUL(int a,int x,int Mod){
// return (INT)a*x%Mod;
int ans=0,w=a;
while(x){
if(x&1)ans=(ans+w)%Mod;
w=(w+w)%Mod;
x>>=1;
}
return ans;
}
const int G1=1145141;
const int MMod=1000000000000000003ll;
#define MN 2000005
#define Ls (x<<1)
#define Rs (x<<1|1)
#define mid ((l+r)>>1)
#define piii pair<int,int>
#define mp make_pair
#define x first
#define y second
int aupd(int a){return (a<MMod)?a:(a-MMod);}
int supd(int a){return (a>=0)?a:(a+MMod);}
int pw1[MN],s1[MN];
int sum[MN];
int n,q,a[MN];
void build(){
for(int i=1;i<=n;++i){
s1[i]=aupd(s1[i]+pw1[a[i]]);
sum[i]=sum[i]+a[i];
int to=i+(i&(-i));
s1[to]=aupd(s1[to]+s1[i]);
sum[to]=sum[to]+sum[i];
}
}
void add(int x,int v1,int v2){
while(x<=n){
s1[x]=aupd(s1[x]+v1);
sum[x]+=v2;x+=x&(-x);
}
}
piii ask(int x){
piii res=mp(0,0);
while(x){
res.x=aupd(res.x+s1[x]);
res.y+=sum[x];
x-=x&(-x);
}
return res;
}
piii qry(int l,int r){
piii R=ask(r),L=ask(l-1);
R.x=supd(R.x-L.x);
R.y-=L.y;
return R;
}
signed main(){
n=read();q=read();
pw1[0]=1;
for(int i=1;i<=1000000;++i){
pw1[i]=MUL(pw1[i-1],G1,MMod);
}
for(int i=1;i<=n;++i){
a[i]=read();
}
build();
for(int i=1;i<=q;++i){
int op=read();
if(!op){
int x=read(),y=read();
add(x,supd(pw1[y]-pw1[a[x]]),y-a[x]);
a[x]=y;
}
else{
int l1=read(),r1=read(),l2=read(),r2=read(),len=r1-l1+1;
piii res1=qry(l1,r1),res2=qry(l2,r2);
int k=res2.y-res1.y;
if(abs(k)%len!=0){puts("NO");continue;}
k/=len;
if(k>=0){
res1.x=MUL(res1.x,pw1[k],MMod);
}
else {
k=-k;
res2.x=MUL(res2.x,pw1[k],MMod);
}
if(res1.x==res2.x)puts("YES");
else puts("NO");
}
}
return 0;
}
E题：序列
思路：
代码：
#include<bits/stdc++.h>
using namespace std;
template<typename T>inline T read(){
T f=0,x=0;char c=getchar();
while(!isdigit(c)) f=c=='-',c=getchar();
while(isdigit(c)) x=x*10+c-48,c=getchar();
return f?-x:x;
}
namespace run{
const int N=5009,mod=998244353;
inline int add(int x,int y){return x+y>=mod?x-mod+y:x+y;}
inline int sub(int x,int y){return x>=y?x-y:x+mod-y;}
inline int qpow(int x,int y){
int ret=1;
while(y){
if(y&1) ret=1LL*x*ret%mod;
x=1LL*x*x%mod,y>>=1;
}
return ret;
}
int fac[N],ifac[N];
inline int C(int n,int m){
if(n<0 || m<0 || n<m) return 0;
return 1LL*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}inline int calc(int n,int d,int m){return sub(C(n,n-d-m),C(n,n-d-m+1));}
int f[N][N],n,k,inv;
int main(){
n=read<int>(),k=read<int>(),inv=qpow(n,mod-2);
fac[0]=ifac[0]=ifac[1]=1;
for(int i=1;i<=n;i++) fac[i]=1LL*fac[i-1]*i%mod;
ifac[n]=qpow(fac[n],mod-2);
for(int i=n-1;i>=1;i--) ifac[i]=1LL*ifac[i+1]*(i+1)%mod;
assert(1LL*fac[n-1]*ifac[n-1]%mod==1);
f[0][0]=1;
for(int i=1;i<=k;i++){
int sum=f[i-1][n];
for(int j=min(n,i);j;j--){
sum=(1LL*sum*j%mod*inv+f[i-1][j-1])%mod;
f[i][j]=1LL*inv*(n-j+1)%mod*sum%mod;
}
}
int chk=0;
for(int i=0;i<=n;i++) chk=add(chk,f[k][i]);
assert(chk==1);
for(int i=1;i<=n;i++) f[k][i]=1LL*f[k][i]*qpow(C(n,i),mod-2)%mod;
int ans=0;
for(int i=1;i<=n;i++)
for(int j=2;j<=n && min(i,n-i)>=j/2;j+=2)
ans=(1LL*f[k][i]*calc(n,i,(j-2*i)/2)%mod*j+ans)%mod;
printf("%d\n",ans);
return 0;
}
}
int main(){
#ifdef my
freopen("sequence.in","r",stdin);
freopen("sequence.out","w",stdout);
#endif
return run::main();
}F题：一次函数
思路：
代码：
#include <bits/stdc++.h>
using namespace std;
const int Maxi = 1030, Maxn = 20005;
int n, m, ct, bloc, f[Maxn][Maxi], pos[Maxi], fac[Maxi];
pair <int, int> from[Maxn][Maxi];
long long d, p, goal, val[Maxn], arr[Maxn];
pair <int, long long> ans[Maxn];
map <int, int> Ma;
typedef pair <long long, long long> pll;
pll G;
pll operator * (const pll &x, const pll &y)
{
return make_pair((x.first * y.second + x.second * y.first) % p, x.second * y.second % p);
}
long long gcd(long long x, long long y)
{
return x == 0 ? y : gcd(y % x, x);
}
long long fast_pow(long long x, long long y)
{
long long ans = 1, now = x;
while (y)
{
if (y & 1) ans = ans * now % p;
now = now * now % p;
y >>= 1;
}
return ans;
}
pll fast_pow(pll x, long long y)
{
pll ans = make_pair(0, 1), now = x;
while (y)
{
if (y & 1) ans = ans * now;
now = now * now;
y >>= 1;
}
return ans;
}
void dp(void)
{
int maxi = (1 << ct) - 1;
memset(f, 0x3f, sizeof(f));
f[0][maxi] = 0;
for (int i = 1; i <= n; i++)
{
int sta = 0;
long long tmp_val = val[i] / gcd(val[i], goal);
for (int j = 1; j <= ct; j++)
if (tmp_val % fac[j] == 0) sta |= 1 << (j - 1);
for (int j = 1; j <= maxi; j++)
if (f[pos[j]][j] + 1 < f[pos[j & sta]][j & sta])
{
pos[j & sta] = i;
f[i][j & sta] = f[pos[j]][j] + 1;
from[i][j & sta] = make_pair(pos[j], j);
}
}
}
int get_unit(void)
{
int maxi = sqrt(p - 1);
for (int i = 2; ; i++)
{
for (int j = 2; j <= maxi; j++)
if ((p - 1) % j == 0)
{
if (fast_pow(i, j) == 1) goto END;
if (fast_pow(i, (p - 1) / j) == 1) goto END;
}
return i;
END:;
}
}
void get_factor(void)
{
int maxi = sqrt(p - 1), tmp = p - 1;
for (int i = 2; i <= maxi; i++)
if (tmp % i == 0)
{
fac[++ct] = i;
while (tmp % i == 0) tmp /= i;
}
if (tmp != 1) fac[++ct] = tmp;
fac[++ct] = p;
}
pll exgcd(long long x, long long y)
{
if (!x)
return make_pair(0, 1);
else
{
pll tmp = exgcd(y % x, x);
return make_pair(tmp.second - tmp.first * (y / x), tmp.first);
}
}
long long mod(__int128 x)
{
return (x % (p * (p - 1)) + (p * (p - 1))) % (p * (p - 1));
}
void work(int lt, long long goal_now)
{
long long g = p * (p - 1), div = gcd(goal_now, p * (p - 1));
for (int i = lt + 1; i <= m; i++)
g = gcd(g, arr[i]);
div = gcd(g, arr[lt]);
arr[lt] /= div, g /= div, goal_now /= div;
pll result = exgcd(arr[lt], g);
ans[lt].second = mod((__int128) result.first * goal_now);
if (lt == m - 1)
{
ans[m].second = mod((__int128) result.second * goal_now);
return ;
}
work(lt + 1, mod(mod((__int128) result.second * goal_now) * (__int128) g));
}
void BSGS_init(void)
{
bloc = sqrt(n * p);
long long now = 1;
d = get_unit();
for (int i = 0; i < bloc; i++)
{
Ma[now] = i;
now = now * d % p;
}
}
long long BSGS(pll val_now)
{
long long inv = fast_pow(fast_pow(d, bloc), p - 2), now_inv = 1;
for (int j = 0; j < p / bloc + 1; j++)
{
if (Ma[now_inv * val_now.second % p])
{
long long tmp = Ma[now_inv * val_now.second % p] + j * (long long) bloc;
pll tmp_val = val_now * fast_pow(make_pair(1, d), p * (p - 1) - tmp);
return ((p - 1) * (p - tmp_val.first * d % p) + tmp) % (p * (p - 1));
}
(now_inv *= inv) %= p;
}
}
int main()
{
scanf("%d%lld%lld%lld", &n, &p, &G.first, &G.second);
BSGS_init();
get_factor();
goal = BSGS(G);
long long g = gcd(p * (p - 1), goal);
for (int i = 1; i <= n; i++)
{
pll x;
scanf("%lld%lld", &x.first, &x.second);
val[i] = BSGS(x), g = gcd(g, val[i]);
}
for (int i = 1; i <= n; i++) val[i] /= g;
dp();
if (f[pos[0]][0] > n)
{
puts("-1");
return 0;
}
pair <int, int> now = make_pair(pos[0], 0);
while (now.first)
{
arr[++m] = val[now.first];
ans[m].first = now.first;
now = from[now.first][now.second];
}
work(1, goal);
printf("%d\n", m);
for (int i = 1; i <= m; i++)
printf("%d %lld\n", ans[i].first, ans[i].second);
return 0;
}希望对大家有所帮助。