本文精选了算法竞赛中的经典题目，包括递归、扩展中国剩余定理、分形图、思维题等，提供了详细的题解思路和代码实现，适合算法学习者深入理解各种算法的应用。
目录
D 、Han Xin and His Troops （扩展CRT）
B 、Coffee Chicken （递归）
题意：
给出两个字符串，按照斐波那契数列进行操作，然后给出一个n和k，计算第n个字符串从第k个开始数数10个并输出，不够就都输出。
分析：
直接dfs即可，还有判断长度1e12后的其实跟前面是一样的。
#include<bits/stdc++.h>
#define mm(a,b) memset(a,b,sizeof(a))
#define ACCELERATE (ios::sync_with_stdio(false),cin.tie(0))
#define pii pair<int,int>
#define pid pair<double,double>
#define pil pair<long long,long long>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define PI acos(-1.0)
#define E exp(1.0)
//#define io
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
ll a[60];
string s,t;
string dfs(int n,ll k,int cnt){
if(n==1){
return s.substr(k-1,min(cnt,(int)s.size()-(int)k+1));
}else if(n==2){
return t.substr(k-1,min(cnt,(int)t.size()-(int)k+1));
}
if(k<a[n-2]&&k+cnt-1<=a[n-2]) return dfs(n-2,k,cnt);
else if(k>a[n-2]) return dfs(n-1,k-a[n-2],cnt);
else return dfs(n-2,k,a[n-2]-k+1)+dfs(n-1,1,cnt-(a[n-2]-k+1));
}
int main(){
#ifdef io
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
a[1]=6;a[2]=7;
int tot=3;
for(;;tot++){
a[tot]=a[tot-1]+a[tot-2];
if(a[tot]>1e12) break;
}
tot++;
a[tot]=a[tot-1]+a[tot-2];
s="COFFEE";
t="CHICKEN";
int T;
scanf("%d",&T);
while(T--){
int n;
ll k;
scanf("%d%lld",&n,&k);
while(n>tot){
n-=2;
}
cout<<dfs(n,k,10)<<endl;
}
return 0;
}
D 、Han Xin and His Troops （扩展CRT）
题意：
很版得题
分析：
非要让我拉个java大数得模板？其实我不想写这个题（在网上抄了一个）。
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static int N = 100010;
public static BigInteger a[] = new BigInteger[N];
public static BigInteger m[] = new BigInteger[N];
public static int k;
public static BigInteger mm,xx,yy;
public static BigInteger exgcd(BigInteger a,BigInteger b){
if(b.compareTo(BigInteger.ZERO)==0){
xx=BigInteger.ONE;
yy=BigInteger.ZERO;
return a;
}
BigInteger z,d=exgcd(b,a.mod(b));
z=xx;
xx=yy;
yy=z.subtract(a.divide(b).multiply(yy));
return d;
}
public static BigInteger china(){
BigInteger c,d,ans=a[1],M=m[1];
for(int i=2;i<=k;i++){
c=(a[i].subtract(ans.mod(m[i]).add(m[i])).mod(m[i]));
d=exgcd(M,m[i]);
if(c.mod(d).compareTo(BigInteger.ZERO)!=0) return new BigInteger("-1");
xx=xx.multiply(c.divide(d)).mod(m[i].divide(d));
ans=ans.add(xx.multiply(M));
M=M.multiply(m[i].divide(d));
ans=ans.mod(M).add(M).mod(M);
}
return ans;
}
public static void main(String[] args) {
Scanner cin = new Scanner(System.in );
k=cin.nextInt(); mm=cin.nextBigInteger();
for(int i=1;i<=k;i++){
m[i]=cin.nextBigInteger();
a[i]=cin.nextBigInteger();
}
BigInteger pm=china(),fu1=new BigInteger("-1");
if(pm.compareTo(fu1)==0) System.out.println("he was definitely lying");
else if(pm.compareTo(mm)>0) System.out.println("he was probably lying");
else System.out.println(pm);
}
}E 、Hilbert Sort （分形图递归）
题意：
给出一个分形图上点的坐标，然后按照路径排序后输出。
分析：
找出每块之间的关系，然后递归计算即可。
#include<bits/stdc++.h>
#define mm(a,b) memset(a,b,sizeof(a))
#define ACCELERATE (ios::sync_with_stdio(false),cin.tie(0))
#define pii pair<int,int>
#define pid pair<double,double>
#define pil pair<long long,long long>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define PI acos(-1.0)
#define E exp(1.0)
//#define io
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
ll dfs(int n, ll x, ll y) {
if(n==0) return 1ll;
ll m=1ll<<(n-1);
if (x<=m&&y<=m){
return dfs(n-1,y,x);
}
if(x<=m&&y>m){
return 3ll*m*m+dfs(n-1,m-(y-m)+1,m-x+1);
}
if(x>m&&y<=m) {
return 1ll*m*m+dfs(n-1,x-m,y);
}
if(x>m&&y>m){
return 2ll*m*m+dfs(n-1,x-m,y-m);
}
}
const int maxn=1e6+10;
struct node{
ll x,y;
ll id;
bool operator < (node p){
return id<p.id;
}
}p[maxn];
int main(){
#ifdef io
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int n,k;
scanf("%d%d",&n,&k);
for(int i=0;i<n;i++){
scanf("%lld%lld",&p[i].x,&p[i].y);
p[i].id=dfs(k,p[i].x,p[i].y);
// cout<<p[i].x<<" "<<p[i].y<<" "<<p[i].id<<endl;
}
sort(p,p+n);
for(int i=0;i<n;i++){
printf("%lld %lld\n",p[i].x,p[i].y);
}
return 0;
}F、 Popping Balloons （思维）
题意：
有n个气球，现在可以选择射破三行气球和三列气球，而且保证相邻行和列间距相同，问最多能射裂多少气球
分析：
求出横着打和竖着打所有结果，按个数从大到小排序，最大值肯定就从前几大的打法中组合出，
而且取出前四大就可以了(举不出反例，但是我太菜不太会证明)，
#include<bits/stdc++.h>
#define mm(a,b) memset(a,b,sizeof(a))
#define ACCELERATE (ios::sync_with_stdio(false),cin.tie(0))
#define pii pair<int,int>
#define pid pair<double,double>
#define pil pair<long long,long long>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define rush() int T;scanf("%d",&T);while(T--)
#define sc(a) scanf("%d",&a)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define PI acos(-1.0)
#define E exp(1.0)
//#define io
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
//#define gc getchar
char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
inline int read(){
int ret=0,f=0;char c=gc();
while(!isdigit(c)){if(c=='-')f=1;c=gc();}
while(isdigit(c)){ret=ret*10+c-48;c=gc();}
if(f)return -ret;return ret;
}
const int maxn=1e5+10;
int fx[maxn*3],fy[maxn*3];
struct node{
int x,y;
bool operator < (const node p) const {
return x>p.x;
}
}a[maxn],w[maxn];
int main(){
#ifdef io
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int n,r;
sc(n),sc(r);
rep(i,1,n){
scanf("%d%d",&a[i].x,&a[i].y);
fx[a[i].x]++;
}
rep(i,0,1e5){
w[i].x=fx[i]+fx[i+r]+fx[i+r+r];
w[i].y=i;
}
sort(w,w+maxn);
int ans=0;
rep(i,0,100){
mm(fy,0);
rep(j,1,n){
if(a[j].x!=w[i].y&&a[j].x!=w[i].y+r&&a[j].x!=w[i].y+r+r){
fy[a[j].y]++;
}
}
int res=0;
rep(j,0,1e5){
res=max(res,fy[j]+fy[j+r]+fy[j+r+r]);
}
ans=max(ans,w[i].x+res);
}
printf("%d\n",ans);
return 0;
}
H、Stammering Chemists （水题）
太简单就不说了。
#include<bits/stdc++.h>
#define mm(a,b) memset(a,b,sizeof(a))
#define ACCELERATE (ios::sync_with_stdio(false),cin.tie(0))
#define pii pair<int,int>
#define pid pair<double,double>
#define pil pair<long long,long long>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define PI acos(-1.0)
#define E exp(1.0)
//#define io
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
int in[10];
int g[10][10];
int main(){
#ifdef io
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int T;
scanf("%d",&T);
while(T--){
mm(in,0);
mm(g,0);
for(int i=1;i<=5;i++){
int u,v;
scanf("%d%d",&u,&v);
in[u]++;
in[v]++;
g[u][v]=1;
g[v][u]=1;
}
int cnt[10]={0};
rep(i,1,6) cnt[in[i]]++;
if(cnt[4]){
printf("2,2-dimethylbutane\n");
}else if(cnt[3]==2){
printf("2,3-dimethylbutane\n");
}else if(cnt[1]==2){
printf("n-hexane\n");
}else{
int tmp;
rep(i,1,6)if(in[i]==3){
tmp=i;break;
}
int one=0;
rep(i,1,6){
if(g[tmp][i]&&in[i]==1) one++;
}
if(one==2){
printf("2-methylpentane\n");
}else{
printf("3-methylpentane\n");
}
}
}
return 0;
}
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