The Preliminary Contest for ICPC China Nanchang National Invitational

Problem A PERFECT NUMBER PROBLEM

https://nanti.jisuanke.com/t/38220

题解：打表

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){//scanf("%d",&n);printf("6\n28\n496\n8128\n33550336\n");//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem B Greedy HOUHOU

https://nanti.jisuanke.com/t/38221

Problem C Angry FFF Party

https://nanti.jisuanke.com/t/38222

题意：f(x)是斐波那契数的第x项，g(x)=f(f(x))，对于给定的n，将其表示成若干个g数的和，要求字典序最小，n<10^100000

题解：矩阵快速幂求一下f(x)，预处理前29项g(x)。

可以发现g(x)的增长速度非常快，其实在第29项就已经超出10^100000，由于后面的项差过大，所以方法基本唯一，所以就把贪心选一下，然后前几项在特判一下取个最小字典序就ok了。
JAVA版本一

import java.io.*;
import java.math.*;
import java.util.*;
public class Main
{static int maxn = 29;static BigInteger []w = new BigInteger[maxn];static int []v = new int[maxn];public static BigInteger f(BigInteger n){n = n.subtract(BigInteger.ONE);BigInteger [][]a = new BigInteger[2][2];BigInteger [][]b = new BigInteger[2][2];BigInteger [][]c = new BigInteger[2][2];for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) a[i][j] = BigInteger.ZERO;a[0][0] = a[1][0] = a[0][1] = BigInteger.ONE; for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) b[i][j] = BigInteger.ZERO;for(int i = 0; i < 2; i++) b[i][i] = BigInteger.ONE;while(n.compareTo(BigInteger.ZERO) > 0){if(n.mod(new BigInteger("2")).equals(BigInteger.ONE)){for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){c[i][j] = BigInteger.ZERO;for(int k = 0; k < 2; k++){c[i][j] = c[i][j].add(b[i][k].multiply(a[k][j]));}}}for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){b[i][j] = c[i][j];}}}for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){c[i][j] = BigInteger.ZERO;for(int k = 0; k < 2; k++){c[i][j] = c[i][j].add(a[i][k].multiply(a[k][j]));}}}for(int i = 0; i < 2; i++){for(int j = 0; j < 2; j++){a[i][j] = c[i][j];}}n = n.divide(new BigInteger("2"));}//System.out.println("len " + b[0][0].toString().length());return b[0][0];}public static void main(String[] args){Scanner cin = new Scanner (System.in);for(int i = 1; i < maxn; i++){w[i] = f(f(BigInteger.valueOf(i)));}// System.out.println(w[6] + " " + w[7]);int T = cin.nextInt();while(T > 0){T--;BigInteger n = cin.nextBigInteger();int tot = 0;for(int i = maxn - 1; i > 0; i--){if(n.equals(new BigInteger("5")) && i >= 4){v[tot++] = 4;v[tot++] = 3;v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("4")) && i >= 4){v[tot++] = 4;v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("3")) && i >= 3){v[tot++] = 3;v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("2")) && i >= 2){v[tot++] = 2;v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.equals(new BigInteger("1")) && i >= 1){v[tot++] = 1;n = BigInteger.ZERO;break;}if(n.compareTo(w[i]) >= 0){v[tot++] = i;n = n.subtract(w[i]);}}if(!n.equals(BigInteger.ZERO)) System.out.println(-1);else{for(int i = tot - 1; i >= 0; i--) System.out.printf("%d%c", v[i], i == 0 ? '\n' : ' ');}}cin.close();}
}

Problem D Match Stick Game

https://nanti.jisuanke.com/t/38223

题解：

先将表达式表示成用最少火柴棍能表达的形式，然后在此基础上添加火柴棍。

先预处理出i位数添加j个火柴棍能够得到的最大值，然后背包一下就ok了。

举个例子

1+1+222

最简表示1-1-111，剩余11根棒子。

一共三项，1，-1，-111，然后dp[i][j]表示前i项加j根棒子的和的最大值，转移显然。

关键是预处理比较麻烦。

#include <bits/stdc++.h>
#define my_max(a, b) ((a) > (b) ? (a) : (b))
#define my_min(a, b) ((a) < (b) ? (a) : (b))
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define rep(i, s, t) for(int i = (int)(s); i <= (int)(t); i++)
#define rev(i, t, s) for(int i = (int)(t); i >= (int)(s); i--)
#define lson rt << 1
#define rson rt << 1 | 1
#define sz(x) (int)(x).size()typedef long long ll;
typedef long double lb;
typedef std::pair<int, int> pii;
typedef std::pair<ll, ll> pll;
typedef std::vector<int> VI;
typedef std::vector<ll> VL;
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> rbtree;
ll gcd(ll x, ll y){ return y % x == 0 ? x : gcd(y % x, x); }
template<class T>T my_abs(T a){ if(a < 0) a = -a; return a; }
inline ll read(){ll ret = 0, sign = 1;char c = getchar();while(c < '0' || c > '9'){ if(c == '-') sign = -1; c = getchar();}while(c >= '0' && c <= '9'){ ret = ret * 10 + c - '0'; c = getchar(); }return ret * sign;
}
inline void write(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) write(x / 10);putchar(x % 10 + '0');
}//void expand_stack_space(){
//    int size = 256 << 20; // 256MB
//    char *p = (char*)malloc(size) + size;
//    __asm__("movl %0, %%esp\n" :: "r"(p));
//}
const int maxn = 100 + 7;
const ll inf = 1e18;
ll dp[maxn][maxn * 10], sa[maxn];
ll a[maxn][maxn], b[maxn][maxn];
int n;
char str[maxn];
int count(char c){if(c == '-' || c == '1') return 0;if(c == '7' || c == '+') return 1;if(c == '4') return 2;if(c == '2' || c == '3' || c == '5') return 3;if(c == '0' || c == '6' || c == '9') return 4;if(c == '8') return 5;
}
char icount(int c){if(c == 1) return '7';if(c == 2) return '4';if(c == 3) return '5';if(c == 4) return '9';if(c == 5) return '8';
}
void init(){for(int i = 1; i <= 10; i++){for(int j = 0; j <= 5 * i; j++){int c = j;string s = "";for(int k = 1; k <= i; k++){if(c >= count('9')) s += '9', c -= count('9');else if(c >= count('7')) s += '7', c -= count('7');else s += '1';}for(int k = i; k >= 1; k--){if(c == 0) break;c += count(s[k - 1]);if(c >= count('8')) s[k - 1] = '8', c -= count('8');else{s[k - 1] = icount(c);c = 0;}}ll res = 0;//cout<<"s " + s<<endl;for(char ss : s) res = res * 10 + ss - '0';a[i][j] = res;b[i][j + 1] = res;}b[i][0] = -b[i][1];}
}
int main(){
#ifndef ONLINE_JUDGE
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGEint T;init();scanf("%d", &T);while(T--){scanf("%d", &n);scanf("%s", str);int cnt = 0;for(int i = 0; i < n; i++) cnt += count(str[i]);int num = 0, tot = 0;for(int i = 0; i < n; i++){if(str[i] == '-' || str[i] == '+'){sa[++tot] = num;num = 0;}else num++;}//printf("cnt %d\n", cnt);sa[++tot] = num;for(int i = 0; i <= tot; i++) for(int j = 0; j <= cnt; j++) dp[i][j] = -inf;//比赛时初始化写错了， 还过了。emmmmdp[0][0] = 0;for(int i = 1; i <= tot; i++){int len = sa[i];//printf("len %d\n", len);if(i == 1){for(int k = 0; k <= 5 * len; k++){for(int j = cnt; j >= k; j--){dp[i][j] = max(dp[i][j], dp[i - 1][j - k] + a[len][k]);}}}else{for(int k = 0; k <= 5 * len; k++){for(int j = cnt; j >= k; j--){dp[i][j] = max(dp[i][j], dp[i - 1][j - k] + b[len][k]);}}}}printf("%lld\n", dp[tot][cnt]);}return 0;
}

Problem E Card Game

https://nanti.jisuanke.com/t/38224

Problem F Information Transmitting

https://nanti.jisuanke.com/t/A2254

Problem G tsy's number

https://nanti.jisuanke.com/t/38226

题解：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;typedef unsigned int ll;const int N = 1e7 + 10;int prime[N], phi[N], tot;
ll f[N], gg[N], fu[N], res[N];
bool is_prime[N];void init() {memset(is_prime, true, sizeof(is_prime));for (ll i = 1; i < N; i++)f[i] = f[i - 1] + 1;for (ll i = 1; i < N; i++) {gg[i] = gg[i - 1] + i;f[i] = f[i] * gg[i];}for (ll i = 1; i < N; i++) {gg[i] = gg[i - 1] + i * i;f[i] = f[i] * gg[i];}phi[1] = gg[1] = fu[1] = 1;for (int i = 2; i < N; i++) {if (is_prime[i]) {prime[tot++] = i;fu[i] = i;phi[i] = i - 1;gg[i] = i - 2;}for (int j = 0; j < tot && i * prime[j] < N; j++) {is_prime[i * prime[j]] = false;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];fu[i * prime[j]] = fu[i] * prime[j];if (fu[i] == i) {gg[i * prime[j]] = phi[i * prime[j]] - phi[i];}else {gg[i * prime[j]] = gg[i / fu[i]] * gg[fu[i * prime[j]]];}break;}else {fu[i * prime[j]] = prime[j];phi[i * prime[j]] = phi[i] * phi[prime[j]];gg[i * prime[j]] = gg[i] * gg[prime[j]];}}}for (ll i = 1; i < N; i++) {res[i] = i * i * i * gg[i] + res[i - 1];}
}int n;void solve() {ll ans = 0;for (ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans += (res[r] - res[l - 1]) * f[n / l];}printf("%d\n", ans % (1 << 30));
}int main() {//freopen("0in.txt", "r", stdin);init();int T;scanf("%d", &T);while (T--) {scanf("%d", &n);solve();}return 0;
}

Problem H Coloring Game

https://nanti.jisuanke.com/t/38227

题解：快速幂+规律

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};ll POW(ll a,ll b,ll c){ll res=1;ll base=a%c;while(b){if(b&1)res=(res*base)%c;base=(base*base)%c;b>>=1;}return res;
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);if(n==1){cout<<1<<endl;return 0;}cout<<(4*POW(3,n-2,MOD))%MOD<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

Problem I Max answer

https://nanti.jisuanke.com/t/38228

题意：求区间最小值×区间和的最大值。

题解：用单调栈维护以每个数为最小值能拓展到最左边和最右边。然后用线段树维护一个前缀和。

对于一个数如果是正数查询当前位置到最右端的前缀和最大值，然后查询最左端到当前点的最小值做减法。

对于一个数如果是负数查询正好和上述情况相反。

C++版本一

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define sca(x) scanf("%d",&x)
#define N 500005
const LL inf = 1e18;int a[N],L[N],R[N];
LL sum[N];
stack<int>S;
struct node
{LL mmax,mmin;
}T[N*4];void build(int rt,int l,int r)
{if(l==r){T[rt].mmax=sum[l];T[rt].mmin=sum[l];return ;}int m=(l+r)>>1;build(rt<<1,l,m);build(rt<<1|1,m+1,r);T[rt].mmax=max(T[rt<<1].mmax,T[rt<<1|1].mmax);T[rt].mmin=min(T[rt<<1].mmin,T[rt<<1|1].mmin);
}LL query_max(int rt,int l,int r,int ql,int qr)
{if(l>=ql && r<=qr){return T[rt].mmax;}int m=(l+r)>>1;LL ans1=-inf,ans2=-inf;if(ql<=m) ans1=query_max(rt<<1,l,m,ql,qr);if(qr>m) ans2=query_max(rt<<1|1,m+1,r,ql,qr);return max(ans1,ans2);
}LL query_min(int rt,int l,int r,int ql,int qr)
{if(l>=ql && r<=qr){return T[rt].mmin;}int m=(l+r)>>1;LL ans1=inf,ans2=inf;if(ql<=m) ans1=query_min(rt<<1,l,m,ql,qr);if(qr>m) ans2=query_min(rt<<1|1,m+1,r,ql,qr);return min(ans1,ans2);
}int main()
{int n;sca(n);sum[0]=0;for(int i=1;i<=n;i++){sca(a[i]);sum[i]=sum[i-1]+a[i];}for(int i=1;i<=n;i++){while(!S.empty()&&a[S.top()]>=a[i])S.pop();if(S.size()==0)L[i]=0;else L[i]=S.top();S.push(i);}while(!S.empty())S.pop();for(int i=n;i>=1;i--){while(!S.empty()&&a[S.top()]>=a[i])S.pop();if(S.size()==0)R[i]=n;else R[i]=S.top()-1;S.push(i);}build(1,0,n);LL ans=-inf;for(int i=1;i<=n;i++){if(a[i]>=0){LL ans1=query_max(1,0,n,i,R[i]);LL ans2=query_min(1,0,n,L[i],i);ans=max(ans,a[i]*(ans1-ans2));}else{LL ans1=query_min(1,0,n,i,R[i]);LL ans2=query_max(1,0,n,L[i],i);ans=max(ans,a[i]*(ans1-ans2));}}cout<<ans<<endl;
}

Problem J Distance on the tree

https://nanti.jisuanke.com/t/38229

题意：给你一棵树，每条边都有一个权值，m个询问，问你从x到y的路径上有多少条边的值<=w。

题解：LCA
一边dfs一边建树，每一个点的状态是从它父亲的状态转移过来的。那么两个点之间的所有可能就是一个点到他们的lca+另一个点到他们的lca。注意存边权的话，是要存在儿子节点上的。

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
int ls[maxn*40],rs[maxn*40],sum[maxn*40],rt[maxn*40];
int tot,a[maxn],b[maxn];
int n,m;
struct node
{int to,next,wei;
}e[maxn<<1];
int head[maxn],fa[maxn][25],cnt,dep[maxn];
void add(int x,int y,int z)
{e[cnt].to=y;e[cnt].next=head[x];e[cnt].wei=z;head[x]=cnt++;
}
void update(int l,int r,int &root,int last,int val)
{root=++tot;ls[root]=ls[last];rs[root]=rs[last];sum[root]=sum[last]+1;if(l==r)return ;int mid=l+r>>1;if(mid>=val)update(l,mid,ls[root],ls[last],val);elseupdate(mid+1,r,rs[root],rs[last],val);
}
int query(int l,int r,int ql,int qr,int num)
{if(l==r)return sum[qr]-sum[ql];int mid=l+r>>1;if(num<=mid)return query(l,mid,ls[ql],ls[qr],num);else{int ans=sum[ls[qr]]-sum[ls[ql]];ans+=query(mid+1,r,rs[ql],rs[qr],num);return ans;}
}
int all;
void dfs(int son,int ffa,int val)
{dep[son]=dep[ffa]+1;fa[son][0]=ffa;if(son!=1)update(1,maxn,rt[son],rt[ffa],val);for(int i=head[son];~i;i=e[i].next){int ne=e[i].to;if(ne==ffa)continue;dfs(ne,son,lower_bound(b+1,b+1+all,e[i].wei)-b);}
}
void deal(){for(int j=1;(1<<j)<=n;j++)for(int i=1;i<=n;i++)fa[i][j]=fa[fa[i][j-1]][j-1];}int lca(int x,int y){if(dep[x]<dep[y])swap(x,y);for(int i=20;i>=0;i--)if(dep[fa[x][i]]>=dep[y])x=fa[x][i];if(x==y)return x;for(int i=20;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];return fa[x][0];
}int main()
{memset(head,-1,sizeof(head));scanf("%d%d",&n,&m);int x,y,z;for(int i=1;i<n;i++){scanf("%d%d%d",&x,&y,&z),b[i]=z;add(x,y,z),add(y,x,z);}sort(b+1,b+n);all=unique(b+1,b+n)-(b+1);tot=0;dfs(1,1,0);deal();while(m--){scanf("%d%d%d",&x,&y,&z);z=upper_bound(b+1,b+1+all,z)-b-1;if(z==0)printf("0\n");else{int l=lca(x,y),ans=0;if(x!=l)ans=query(1,maxn,rt[l],rt[x],z);if(y!=l)ans+=query(1,maxn,rt[l],rt[y],z);printf("%d\n",ans);}}return 0;
}

Problem K MORE XOR

https://nanti.jisuanke.com/t/38230

题解：规律

C++版本一

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define INF 0x3f3f3f3f
#define FI first
#define SE second
#define MP make_pair
#define PI pair<int,int>
#define lson l,m,rt<<1,ls,rs
#define rson m+1,r,rt<<1|1,ls,rs
#define test printf("here!!!\n")
using namespace std;
const int mx=1e5+10;
int n,m,p,q;
int qz[4][4][mx];
int hz[4][4][mx];
int a[mx];
int main()
{int t;scanf("%d",&t);while (t--){memset(qz,0,sizeof(qz));memset(hz,0,sizeof(hz));scanf("%d",&n);for (int i=1;i<=n;++i){scanf("%d",&a[i]);}for (int i=1;i<=n;++i){for (int j=0;j<4;++j){int id=(i-1)%4;int hgf=(1+j)%4;if (id==j){qz[1][j][i]=qz[1][j][i-1]^a[i];}else qz[1][j][i]=qz[1][j][i-1];if (id==j||id==hgf){qz[2][j][i]=qz[2][j][i-1]^a[i];}else qz[2][j][i]=qz[2][j][i-1];if (id==hgf){qz[3][j][i]=qz[3][j][i-1]^a[i];}else qz[3][j][i]=qz[3][j][i-1];}}for (int i=n;i>=1;--i){for (int j=0;j<4;++j){int id=(i-1)%4;int hgf=(1+j)%4;if (id==j){hz[1][j][i]=hz[1][j][i+1]^a[i];}else hz[1][j][i]=hz[1][j][i+1];if (id==j||id==hgf){hz[2][j][i]=hz[2][j][i+1]^a[i];}else hz[2][j][i]=hz[2][j][i+1];if (id==hgf){hz[3][j][i]=hz[3][j][i+1]^a[i];}else hz[3][j][i]=hz[3][j][i+1];}}scanf("%d",&q);int l,r;for (int i=1;i<=q;++i){scanf("%d%d",&l,&r);int dis=(r-l+1)%4;int cxk=(l-1)%4;if (dis!=0) printf("%d\n",qz[dis][cxk][n]^qz[dis][cxk][l-1]^hz[dis][cxk][r+1]);else printf("0\n");}}
}

C++版本二

题解：规律

打表代码：

    while(1){scanf("%d",&n);for(int l=1;l<=n;l++){for(int r=l;r<=n;r++){memset(vis,0,sizeof(vis));for(int l1=l;l1<=r;l1++){for(int r1=l1;r1<=r;r1++){for(int l2=l1;l2<=r1;l2++){for(int r2=l2;r2<=r1;r2++){for(int i=l2;i<=r2;i++){vis[i]=!vis[i];}}}}}cout<<l<<" "<<r<<endl;for(int i=1;i<=12;i++){cout<<vis[i];}cout<<endl;}}}

1、可以发现当区间确定时，是否取用成一个固定的规律；

2、同长度的区间，取法相同；

3、对于一定的区间取法循环节长度为4，例如【1,9】循环节1000，整个区间取法100010001，并且区间长度循环节也为4，例如长度为1的区间和长度为5的区间取法循环节相同；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
int vis[N];
char str;
int qz[4][4][N];//第一维:取法，第二维:开始位置
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}memset(qz,0,sizeof(qz));for(int i=1;i<=n;i++){for(int j=0;j<4;j++){int pos=(i-1)%4;int dis=(j+1)%4;if(pos==j){qz[1][j][i]=qz[1][j][i-1]^a[i];}else{qz[1][j][i]=qz[1][j][i-1];}if(pos==j||dis==pos){qz[2][j][i]=qz[2][j][i-1]^a[i];}else{qz[2][j][i]=qz[2][j][i-1];}if(dis==pos){qz[3][j][i]=qz[3][j][i-1]^a[i];}else{qz[3][j][i]=qz[3][j][i-1];}}}scanf("%d",&m);while(m--){scanf("%d%d",&l,&r);int pos=(r-l+1)%4;int dis=(l-1)%4;int res=qz[pos][dis][r]^qz[pos][dis][l-1];cout<<res<<endl;}}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

C++版本三

Problem L qiqi'tree

https://nanti.jisuanke.com/t/38231

Problem M Subsequence

https://nanti.jisuanke.com/t/38232

题解：预处理+贪心

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str[N],s[N];
int pos[N][130];
struct node{};
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){cin>>str+1;scanf("%d",&n);int len=strlen(str+1);for(int i=0;i<128;i++){pos[len][i]=len+1;}for(int i=len-1;i>=0;i--){for(int j=0;j<128;j++){pos[i][j]=pos[i+1][j];}pos[i][str[i+1]]=i+1;}while(n--){scanf("%s",s);int l=strlen(s);int id=pos[0][s[0]];int r=0;while(id<=len&&r<l){r++;id=pos[id][s[r]];}//cout<<l<<" "<<r<<endl;cout<<((r==l)?"YES":"NO")<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}