題解：
https://files.cnblogs.com/files/clrs97/2018CCPC%E5%A5%B3%E7%94%9F%E8%B5%9Bsol.pdf
Code：
1001. CCPC直播
#include<cstdio>
#include<cstring>
int Case,n,i,x;char s[100];
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%s",s);
n=strlen(s);
for(i=0;i<3-n;i++)putchar(' ');
printf("%s|",s);
scanf("%s",s);
n=strlen(s);
printf("%s",s);
for(i=0;i<16-n;i++)putchar(' ');
scanf("%s",s);
printf("|%s|[",s);
scanf("%s",s);
n=strlen(s);
if(s[0]=='R'&&s[1]=='u'){
scanf("%d",&x);
for(i=0;i<x;i++)putchar('X');
for(i=0;i<10-x;i++)putchar(' ');
}else if(s[0]=='F'){
printf(" AC* ");
}else{
printf(" %s",s);
for(i=0;i<6-n;i++)putchar(' ');
}
puts("]");
}
}
1002. 口算訓練
#include<cstdio>
#include<vector>
using namespace std;
const int N=100010;
int Case,n,m,i,x,L,R;vector<int>v[N];
inline void add(int n,int x){
for(int i=2;i*i<=n;i++)while(n%i==0)v[i].push_back(x),n/=i;
if(n>1)v[n].push_back(x);
}
inline bool ask(int x,int k){
int l=0,r=v[x].size()-1,t=-1,mid;
while(l<=r)if(v[x][mid=(l+r)>>1]>=L)r=(t=mid)-1;else l=mid+1;
if(t<0)return 0;
r=v[x].size()-1;
while(k--){
if(t>r)return 0;
if(v[x][t]>R)return 0;
t++;
}
return 1;
}
inline bool solve(int n){
for(int i=2;i*i<=n;i++)if(n%i==0){
int t=0;
while(n%i==0)n/=i,t++;
if(!ask(i,t))return 0;
}
if(n>1)return ask(n,1);
return 1;
}
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d%d",&n,&m);
for(i=2;i<N;i++)v[i].clear();
for(i=1;i<=n;i++)scanf("%d",&x),add(x,i);
while(m--)scanf("%d%d%d",&L,&R,&x),puts(solve(x)?"Yes":"No");
}
}
1003. 缺失的資料範圍
#include<cstdio>
typedef long long ll;
int Case,a,b;ll K,l,r,mid,ans;
inline ll mul(ll a,ll b){
if(!a||!b)return 0;
if(a>(K+5)/b)return K+5;
a*=b;
return a>K+5?K+5:a;
}
inline ll getpow(ll a,int b){
ll t=1;
while(b--)t=mul(t,a);
return t;
}
inline int getlog(ll n){
ll o=n;
while(o%2==0)o/=2;
int t=0;
while(n>1)n/=2,t++;
if(o>1)t++;
return t;
}
inline ll cal(ll n,int a,int b){
return mul(getpow(n,a),getpow(getlog(n),b));
}
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d%d%lld",&a,&b,&K);
l=2,r=K,ans=1;
while(l<=r){
mid=(l+r)>>1;
if(cal(mid,a,b)<=K)l=(ans=mid)+1;else r=mid-1;
}
printf("%lld\n",ans);
}
}
1004. 尋寶遊戲
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=55,M=25;
int Case,n,m,K,i,j,k,x,y,z,a[N][N],f[N][N][M][M],ans,s[N],cnt;//in out
inline bool cmp(int x,int y){return x>y;}
inline void up(int&a,int b){if(a<b)a=b;}
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d%d%d",&n,&m,&K);
for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%d",&a[i][j]);
for(i=1;i<=n;i++)for(j=1;j<=m;j++)for(x=0;x<=K;x++)for(y=0;y<=K;y++)f[i][j][x][y]=-1;
up(f[1][1][0][0],a[1][1]);
up(f[1][1][0][1],0);
for(i=1;i<=n;i++)for(j=1;j<=m;j++){
if(i<n){
cnt=0;
for(k=1;k<=K;k++)s[k]=0;
for(k=j+1;k<=m;k++)s[++cnt]=a[i][k];
for(k=1;k<j;k++)s[++cnt]=a[i+1][k];
sort(s+1,s+cnt+1,cmp);
for(k=1;k<=K;k++)s[k]+=s[k-1];
}
for(x=0;x<=K;x++)for(y=0;y<=K;y++)if(~f[i][j][x][y]){
z=f[i][j][x][y];
//move right
if(j<m){
//count in
up(f[i][j+1][x][y],z+a[i][j+1]);
//swap out
if(y<K)up(f[i][j+1][x][y+1],z);
}
//move down
if(i<n)for(k=0;x+k<=K;k++){
//count in
up(f[i+1][j][x+k][y],z+s[k]+a[i+1][j]);
//swap out
if(y<K)up(f[i+1][j][x+k][y+1],z+s[k]);
}
}
}
for(ans=x=0;x<=K;x++)up(ans,f[n][m][x][x]);
printf("%d\n",ans);
}
}
1005. 奢侈的旅行
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<ll,int>P;
const int N=100010,M=200010;
const ll inf=1LL<<60;
int Case,n,m,i,g[N],v[M],a[M],b[M],nxt[M],ed;ll d[N];
priority_queue<P,vector<P>,greater<P> >q;
inline void add(int x,int y,int A,int B){v[++ed]=y;a[ed]=A;b[ed]=B;nxt[ed]=g[x];g[x]=ed;}
int getlog(ll x){
int t=0;
while(x>1)x/=2,t++;
return t;
}
inline void ext(int x,ll y){if(d[x]>y)q.push(P(d[x]=y,x));}
inline bool check(ll lv,ll a,ll b){return a/lv>=((1LL<<b)-1);}
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d%d",&n,&m);
for(ed=0,i=1;i<=n;i++)g[i]=0;
while(m--){
int x,y,a,b;
scanf("%d%d%d%d",&x,&y,&a,&b);
add(x,y,a,b);
}
for(i=1;i<=n;i++)d[i]=inf;
ext(1,1);
while(!q.empty()){
P t=q.top();q.pop();
if(d[t.second]<t.first)continue;
for(i=g[t.second];i;i=nxt[i])if(check(t.first,a[i],b[i]))ext(v[i],t.first+a[i]);
}
if(d[n]==inf)puts("-1");else printf("%d\n",getlog(d[n]));
}
}
1006. 對稱數
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
const int N=200010,M=N*20;
int Case,n,m,i,x,y,z,a[N],g[N],v[N<<1],nxt[N<<1],ed;
int size[N],son[N],d[N],f[N],top[N],tot,root[N],l[M],r[M];
ll sum[M],ran[N],pre[N];
inline void add(int x,int y){v[++ed]=y;nxt[ed]=g[x];g[x]=ed;}
int ins(int x,int a,int b,int c,ll d){
int y=++tot;
sum[y]=sum[x]^d;
if(a==b)return y;
int mid=(a+b)>>1;
if(c<=mid){
l[y]=ins(l[x],a,mid,c,d);
r[y]=r[x];
}else{
l[y]=l[x];
r[y]=ins(r[x],mid+1,b,c,d);
}
return y;
}
void dfs(int x,int y){
f[x]=y;d[x]=d[y]+1;
size[x]=1;son[x]=0;
root[x]=ins(root[y],1,N,a[x],ran[a[x]]);
for(int i=g[x];i;i=nxt[i])if(v[i]!=y){
dfs(v[i],x);
size[x]+=size[v[i]];
if(size[v[i]]>size[son[x]])son[x]=v[i];
}
}
void dfs2(int x,int y){
top[x]=y;
if(son[x])dfs2(son[x],y);
for(int i=g[x];i;i=nxt[i])if(v[i]!=f[x]&&v[i]!=son[x])dfs2(v[i],v[i]);
}
inline int lca(int x,int y){
while(top[x]!=top[y]){
if(d[top[x]]<d[top[y]])swap(x,y);
x=f[top[x]];
}
return d[x]<d[y]?x:y;
}
inline int ask(int A,int B,int C,int D){
int a=1,b=N;
while(a<b){
int mid=(a+b)>>1;
if((sum[l[A]]^sum[l[B]]^sum[l[C]]^sum[l[D]])!=(pre[mid]^pre[a-1])){
b=mid;
A=l[A];
B=l[B];
C=l[C];
D=l[D];
}else{
a=mid+1;
A=r[A];
B=r[B];
C=r[C];
D=r[D];
}
}
return a;
}
int main(){
for(i=1;i<N;i++)ran[i]=ran[i-1]*233+17;
for(i=1;i<N;i++)pre[i]=pre[i-1]^ran[i];
scanf("%d",&Case);
while(Case--){
scanf("%d%d",&n,&m);
for(tot=ed=0,i=1;i<=n;i++)g[i]=0;
for(i=1;i<=n;i++)scanf("%d",&a[i]);
for(i=1;i<n;i++)scanf("%d%d",&x,&y),add(x,y),add(y,x);
dfs(1,0);
dfs2(1,1);
while(m--){
scanf("%d%d",&x,&y);
z=lca(x,y);
printf("%d\n",ask(root[x],root[y],root[z],root[f[z]]));
}
}
}
1007. 賽題分析
#include<cstdio>
#include<algorithm>
using namespace std;
int T,x,y,n,m,i,A,B;
int main(){
scanf("%d",&T);
for(x=1;x<=T;x++){
scanf("%d%d",&n,&m);
A=B=~0U>>1;
for(i=1;i<=n;i++){
scanf("%d",&y);
A=min(A,y);
}
for(i=1;i<=m;i++){
scanf("%d",&y);
B=min(B,y);
}
printf("Problem %d:\n",x+1000);
printf("Shortest judge solution: %d bytes.\n",A);
if(m)printf("Shortest team solution: %d bytes.\n",B);
else puts("Shortest team solution: N/A bytes.");
}
}
1008. quailty演算法
#include<cstdio>
const int N=300010,inf=~0U>>1;
int Case,n,i,a[N],b[N];long long ans;
void solve(int o,int l,int r){
if(o<0||l>=r)return;
int i,j,L=l-1,R=r+1;
for(i=l;i<=r;i++)if(a[i]>>o&1)b[++L]=a[i];else b[--R]=a[i];
for(i=l;i<=r;i++)a[i]=b[i];
solve(o-1,l,L);
solve(o-1,R,r);
int cl=L-l+1,cr=r-R+1;
if(!cl||!cr)return;
if(cl>=3&&cr>=3)return;
int k=1,A=inf,B=inf;
if(cl<3&&cr<3)k++;
for(i=l;i<=L;i++)for(j=R;j<=r;j++){
int t=a[i]^a[j];
if(t<A)B=A,A=t;
else if(t<B)B=t;
}
while(k--){
if(A==inf)return;
ans+=A;
A=B;
B=inf;
}
}
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d",&a[i]);
ans=0;
solve(30,1,n);
printf("%lld\n",ans);
}
}
1009. SA-IS字尾陣列
#include<cstdio>
const int N=1000010;
int Case,n,i;char a[N],f[N];
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d%s",&n,a+1);
f[n]='>';
for(i=n-1;i;i--){
if(a[i]<a[i+1])f[i]='<';
else if(a[i]>a[i+1])f[i]='>';
else f[i]=f[i+1];
}
for(i=1;i<n;i++)putchar(f[i]);
puts("");
}
}
1010. 迴文樹
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
typedef pair<int,int>P;
const int N=100010;
int Case,n,i,j,x,y,a[N],ans;vector<P>g[N];
void dfs(int x,int fx,int y,int fy){
if(a[x]!=a[y])return;
if(fx&&x==y)return;
if(x<=y)ans++;
for(int i=0,j=0,k=0;i<g[x].size();i++)if(g[x][i].second!=fx){
while(j<g[y].size()&&g[y][j].first<g[x][i].first)j++;
while(k<g[y].size()&&g[y][k].first<=g[x][i].first)k++;
for(int o=j;o<k;o++)if(g[y][o].second!=fy)dfs(g[x][i].second,x,g[y][o].second,y);
}
}
int main(){
scanf("%d",&Case);
while(Case--){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d",&a[i]),g[i].clear();
for(i=1;i<n;i++){
scanf("%d%d",&x,&y);
g[x].push_back(P(a[y],y));
g[y].push_back(P(a[x],x));
}
for(i=1;i<=n;i++)sort(g[i].begin(),g[i].end());
ans=0;
for(i=1;i<=n;i++)dfs(i,0,i,0);
for(i=1;i<=n;i++)for(j=0;j<g[i].size();j++)dfs(i,g[i][j].second,g[i][j].second,i);
printf("%d\n",ans);
}
}
1011. 程式碼派對
#include<cstdio>
typedef long long ll;
const int N=100010,M=1010;
int Case,n,m,i,j,e[N][4],s[M][M];ll C3[N],ans;
ll solve(int dx,int dy){
for(i=1;i<=m;i++)for(j=1;j<=m;j++)s[i][j]=0;
for(i=1;i<=n;i++){
int xl=e[i][0]+dx,yl=e[i][1]+dy,xr=e[i][2],yr=e[i][3];
if(xl>xr||yl>yr)continue;
s[xl][yl]++;
s[xr+1][yl]--;
s[xl][yr+1]--;
s[xr+1][yr+1]++;
}
ll ret=0;
for(i=1;i<=m;i++)for(j=1;j<=m;j++){
s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1];
ret+=C3[s[i][j]];
}
return ret;
}
int main(){
for(i=1;i<N;i++)C3[i]=1LL*i*(i-1)*(i-2)/6;
scanf("%d",&Case);
m=1000;
while(Case--){
scanf("%d",&n);
for(i=1;i<=n;i++)for(j=0;j<4;j++)scanf("%d",&e[i][j]);
ans=solve(0,0)-solve(1,0)-solve(0,1)+solve(1,1);
printf("%lld\n",ans);
}
}