A - ArcSoft's Office Rearrangement
均分石子。
好像怎么分答案都一样，于是模拟一遍。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}
template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}
ll a[100005],n,k,x;
ll s;
int main(){
//freopen("in.txt","r",stdin);
int T,cas=1;
cin>>T;
while(T--){
read(n,k);
s=0;
for(int i=1;i<=n;i++)
read(a[i]),s+=a[i];
if(s%k!=0){
printf("Case #%d: -1
",cas++);
continue;
}
ll cmp = s / k;
ll ans = 0;
for(int i=1;i<=n;i++){
if(a[i]==cmp) continue;
if(a[i]<cmp){
if(a[i]+a[i+1]<=cmp){
a[i+1]=a[i]+a[i+1];
ans++;
} else {
a[i+1]=a[i+1]+a[i]-cmp;
ans+=2;
}
} else {
ll tmp = a[i] / cmp;
ll tt = a[i] % cmp;
if(tt!=0){
ans+=2;
a[i+1]+=tt;
}
ans+=tmp-1;
}
}
printf("Case #%d: ",cas++);
cout<<ans<<endl;
}
return 0;
}
B - Bomb
Tarjan之后算一下不同scc里面的cost。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct infonode {
ll x,y,r,d;
}info[2005];
struct tarjan {
const static int maxn = 2e4+7;
const static int maxm = 2e6+7;
struct Edge {
int to,nxt;
Edge(){}
Edge(int _to,int _nxt):to(_to),nxt(_nxt){}
}edge[maxm];
int head[maxn],tot,n;
int Low[maxn],DFN[maxn],Stack[maxn],Belong[maxn];
int Index,top;
int scc;
int num[maxn];
bool Instack[maxn];
int Min[maxn];
inline void init(int n) {
tot = 0;this->n=n;
memset(head,-1,sizeof head);
}
inline void addedge(int u,int v) {
edge[tot]=Edge(v,head[u]);
head[u]=tot++;
}
void Tarjan(int u) {
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for(int i=head[u];i!=-1;i=edge[i].nxt){
v = edge[i].to;
if(!DFN[v]){
Tarjan(v);
Low[u]=min(Low[u],Low[v]);
} else if(Instack[v] && Low[u]>DFN[v]){
Low[u]=DFN[v];
}
}
if(Low[u]==DFN[u]){
scc++;
do{
v=Stack[--top];
Instack[v]=false;
Belong[v]=scc;
num[scc]++;
}while(v!=u);
}
}
int indeg[maxn];
inline ll solve(int n){
memset(DFN,0,sizeof DFN);
memset(num,0,sizeof num);
memset(Instack,0,sizeof Instack);
Index = scc = top = 0;
for(int i=1;i<=n;i++)
if(!DFN[i]) Tarjan(i);
memset(indeg,0,sizeof indeg);
//if(scc==1) return 0;
for(int u=1;u<=n;u++)
for(int i=head[u];i!=-1;i=edge[i].nxt){
int v=edge[i].to;
if(Belong[u]!=Belong[v]){
indeg[Belong[v]]++;
}
}
memset(Min,63,sizeof Min);
for(int i=1;i<=n;i++)
Min[Belong[i]]=min((ll)Min[Belong[i]],info[i].d);
ll ret = 0;
for(int i=1;i<=scc;i++)
if(indeg[i]==0)
ret += (ll)Min[i];
return ret;
}
} g;
int n,cas=1;
template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}
template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}
int main(){
//freopen("in.txt","r",stdin);
int T;
read(T);
while(T--){
read(n);
g.init(n);
for(int i=1;i<=n;i++)
read(info[i].x,info[i].y,info[i].r,info[i].d);
for(int i=1;i<n;i++)
for(int j=i+1;j<=n;j++){
ll dist = (ll)(info[i].x-info[j].x)*(info[i].x-info[j].x)+(info[i].y-info[j].y)*(info[i].y-info[j].y);
if(dist<=info[i].r*info[i].r){
g.addedge(i,j);
}
if(dist<=info[j].r*info[j].r){
g.addedge(j,i);
}
}
ll ans = g.solve(n);
printf("Case #%d: ",cas++);
cout<<ans<<endl;
}
return 0;
}
C - Car
模拟分数，否则卡精度，因为涉及多次除被除除。。。
#include <bits/stdc++.h>
#define maxn 100050
using namespace std;
typedef long long LL;
int t,n;
int a[maxn];
int Case=1;
int main(){
//freopen("in.txt","r",stdin);
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
LL fenzi=a[n]-a[n-1],fenmu=1;
LL time=1;
for(int i=n-1;i>=1;i--){
LL d=a[i]-a[i-1];
fenmu*=d;
LL tmp=fenmu/fenzi;
if(fenmu%fenzi!=0) tmp++;
time+=tmp;
fenzi=d;
fenmu=tmp;
}
printf("Case #%d: %lld
",Case++,time);
}
return 0;
}
D - Difference
每个k预处理一半，然后用中途相遇法，就是类似尺取。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[10][100005],b[10][100005];
ll base[11];
ll P[11][11];
template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}
template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}
void init(){
base[0]=1;
for(int i=1;i<=10;i++)
base[i]=base[i-1]*10ll;
for(int i=0;i<=9;i++){
P[i][1]=i;
for(int j=2;j<=9;j++)
P[i][j]=P[i][j-1]*i;
}
for(int kk=1;kk<=9;kk++){
for(int i=0;i<=99999;i++){
a[kk][i]=P[i%10][kk]+P[i%100/10][kk]+P[i%1000/100][kk]+P[i%10000/1000][kk]
+P[i%100000/10000][kk]-(ll)i*100000ll;
b[kk][i]=P[i%10][kk]+P[i%100/10][kk]+P[i%1000/100][kk]+P[i%10000/1000][kk]
+P[i%100000/10000][kk]-i;
}
}
for(int i=1;i<=9;i++){
sort(a[i],a[i]+100000);
sort(b[i],b[i]+100000);
}
}
int main(){
//freopen("in.txt","r",stdin);
init();
int T,k,cas=1;
ll x;
cin>>T;
while(T--){
read(x,k);
ll res = 0;
for(int l=0,r=99999;l<=99999 && r;){
if(a[k][l]+b[k][r]>x)
r--;
else if(a[k][l]+b[k][r]<x)
l++;
else {
ll l1=0,r1=0;
ll t1=a[k][l];
ll t2=b[k][r];
while(a[k][l]==t1 && l<=99999) l++,l1++;
while(b[k][r]==t2 && r!=0) r--,r1++;
res += l1*r1;
}
}
if(x==0) res--;
printf("Case #%d: ",cas++);
cout<<res<<endl;
}
return 0;
}
F - Four Operations
贪心+枚举。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int cas=1;
char s[55];
int main(){
//freopen("in.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--){
scanf("%s",s+1);
int len=strlen(s+1);
ll ans = -INT_MAX;
for(int i=1;i<=len-4;i++){
for(int j=i+1;j<=len-3;j++){
ll A=0,B=0,C=0,D=0,E=0;
for(int c=1;c<=i;c++)
A=A*10ll+s[c]-'0';
for(int c=i+1;c<=j;c++)
B=B*10ll+s[c]-'0';
C=s[j+1]-'0';
D=s[j+2]-'0';
for(int c=j+3;c<=len;c++)
E=E*10ll+s[c]-'0';
ll tmp = A + B - C * D / E;
ans = max(ans,tmp);
}
}
printf("Case #%d: ",cas++);
cout<<ans<<endl;
}
return 0;
}
K - Kingdom of Obsession
二分图匹配，建立匹配边就行。
若有交合部分，s<n
实际上可以由(1...n)-(s+1...s+n)变成(1...s)-(n+1,...,n+s)，中间部分自己匹配。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1005;
int matches[maxn];
int link[maxn][maxn];
bool used[maxn];
int n,s;
bool find(int x){
for(int j=1;j<=n;j++){
if(link[x][j] && used[j]==false){
used[j]=1;
if(matches[j]==0 || find(matches[j])){
matches[j]=x;
return 1;
}
}
}
return 0;
}
int main(){
//freopen("in.txt","r",stdin);
int T,cas=1;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&s);
if(s==0){
printf("Case #%d: Yes
",cas++);
continue;
}
if(n>s) swap(n,s);
if(n>1000) printf("Case #%d: No
",cas++);
else {
memset(link,0,sizeof link);
memset(matches,0,sizeof matches);
for(int i=s+1;i<=s+n;i++){
for(int j=1;j<=n;j++)
if(i%j==0)
link[j][i-s]=1;
}
int res = 0;
for(int i=1;i<=n;i++){
memset(used,0,sizeof used);
if(find(i)) res++;
}
if(res==n){
printf("Case #%d: Yes
",cas++);
} else {
printf("Case #%d: No
",cas++);
}
}
}
return 0;
}