这里是除去Beginners Problems后的部分
1020 - A Childhood Game
巴什博奕（Bash Game）
#include<bits/stdc++.h>
using namespace std;
int main(void)
{
int t,Case=0;
int n;
char s[10];
scanf("%d",&t);
while(t--)
{
scanf("%d%s",&n,&s);
printf("Case %d: ",++Case);
if(strcmp(s,"Alice")==0)
{
if(n%3==1)
puts("Bob");
else
puts("Alice");
}
else
{
if(n%3==0)
puts("Alice");
else
puts("Bob");
}
}
return 0;
}
1078 - Integer Divisibility
#include<bits/stdc++.h>
using namespace std;
int main(void)
{
int t,Case=0;
int n,digit;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&digit);
int cnt=1;
int tmp=digit;
while(tmp%n!=0)
{
tmp=(tmp*10+digit)%n;
cnt++;
}
printf("Case %d: %d\n",++Case,cnt);
}
return 0;
}
1148 - Mad Counting
#include<bits/stdc++.h>
using namespace std;
map<int,int> ma;
map<int,int>::iterator ite;
int main(void)
{
int t,Case=0;
int n,x;
scanf("%d",&t);
while(t--)
{
ma.clear();
scanf("%d",&n);
while(n--)
{
scanf("%d",&x);
ma[x]++;
}
int ans=0;
for(ite=ma.begin(); ite!=ma.end(); ite++)
{
ans+=(ite->second+ite->first)/(ite->first+1)*(ite->first+1);
}
printf("Case %d: %d\n",++Case,ans);
}
return 0;
}
1179 - Josephus Problem
约瑟夫环
#include<bits/stdc++.h>
using namespace std;
int main(void)
{
int t,Case=0;
int n,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
int ans=0;
for(int i=2;i<=n;i++)
{
ans=(ans+k%i)%i;
}
printf("Case %d: %d\n",++Case,ans+1);
}
return 0;
}
1275 - Internet Service Providers
方程极值
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-6;
int main(void)
{
int t,Case=0;
double n,c;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&n,&c);
printf("Case %d: %d\n",++Case,n>eps?(int)(c/2/n+0.5-eps):0);
}
return 0;
}
1297 - Largest Box
方程极值
#include<bits/stdc++.h>
using namespace std;
double maxx(double w,double l)
{
double a=12;
double b=-4*(w+l);
double c=w*l;
double x=-b-sqrt(b*b-4*a*c);
x/=2*a;
return x;
}
double fun(double w,double l)
{
double x=maxx(w,l);
return (w-2*x)*(l-2*x)*x;
}
int main(void)
{
int t,Case=0;
double w,l;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&w,&l);
printf("Case %d: %f\n",++Case,fun(w,l));
}
return 0;
}
1323 - Billiard Balls
碰撞可以为穿透，那么可以无视别的球直接求最终位置
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x;
int y;
int dx;
int dy;
} p[1010];
int cmp(const node a,const node b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main(void)
{
int t,Case=0;
int l,w,n,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d%d",&l,&w,&n,&k);
char s[5];
for(int i=0; i<n; i++)
{
scanf("%d%d%s",&p[i].x,&p[i].y,s);
if(s[0]=='N')
p[i].dy=1;
else
p[i].dy=-1;
if(s[1]=='W')
p[i].dx=-1;
else
p[i].dx=1;
}
for(int i=0; i<n; i++)
{
p[i].x+=p[i].dx*k;
p[i].y+=p[i].dy*k;
while(!((p[i].x>=0)&&(p[i].x<=l)&&(p[i].y>=0)&&p[i].y<=w))
{
if(p[i].x<0)
p[i].x=-p[i].x;
else if(p[i].x>l)
p[i].x=2*l-p[i].x;
if(p[i].y<0)
p[i].y=-p[i].y;
else if(p[i].y>w)
p[i].y=2*w-p[i].y;
}
}
sort(p,p+n,cmp);
printf("Case %d:\n",++Case);
for(int i=0; i<n; i++)
printf("%d %d\n",p[i].x,p[i].y);
}
return 0;
}
1349 - Aladdin and the Optimal Invitation
求中位数,每个点的人数即访问次数
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x;
int y;
int z;
} p[50050];
int cmp1(const node a,const node b)
{
return a.x<b.x;
}
int cmp2(const node a,const node b)
{
return a.y<b.y;
}
int main(void)
{
int t,Case=0;
int m,n,q;
int x,y,z;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&m,&n,&q);
int sum=0;
for(int i=0; i<q; i++)
{
scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);
sum+=p[i].z;
}
sum=(sum+1)/2;
sort(p,p+q,cmp1);
z=0;
for(int i=0; i<q; i++)
{
z+=p[i].z;
if(z>=sum)
{
x=p[i].x;
break;
}
}
sort(p,p+q,cmp2);
z=0;
for(int i=0; i<q; i++)
{
z+=p[i].z;
if(z>=sum)
{
y=p[i].y;
break;
}
}
printf("Case %d: %d %d\n",++Case,x,y);
}
return 0;
}
1369 - Answering Queries
理解代码块并优化
#include<bits/stdc++.h>
using namespace std;
long long a[100010];
int main(void)
{
int t,Case=0;
int n,q;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&q);
for(int i=0; i<n; i++)
scanf("%lld",&a[i]);
long long sum=0;
for(int i=0; i<n; i++)
sum+=a[i]*(n-1-2*i);
printf("Case %d:\n",++Case);
int index;
long long x,y;
while(q--)
{
scanf("%d",&index);
if(index==1)
printf("%lld\n",sum);
else
{
scanf("%lld%lld",&x,&y);
sum-=a[x]*(n-1-2*x);
a[x]=y;
sum+=a[x]*(n-1-2*x);
}
}
}
return 0;
}
1410 - Consistent Verdicts
问方案数，干脆直接统计有多少个不同的距离
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x;
int y;
} p[701];
long long d[700*700+5];
int main(void)
{
int t,Case=0;
int n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int len=0;
for(int i=0; i<n; i++)
{
scanf("%lld%lld",&p[i].x,&p[i].y);
for(int j=0; j<i; j++)
d[len++]=((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
}
sort(d,d+len);
int ans=unique(d,d+len)-d;
printf("Case %d: %d\n",++Case,ans+1);
}
return 0;
}
1430 - A Question of Time
这题一共交了20发，一共才24人AC，网上找都找不到题解，已刷吐
原本是用那个fun()直接求的，样例和自测数据都能过，交上去无限WA，所以改为二分（如果直接暴力复杂度是12*60*60*13*T==1e10会超时）
#include <bits/stdc++.h>
using namespace std;
//fun()=3600*(2*h0-hx)+60*(2*m0-13*mx)+2*s0
//13*sx=fun()
//13*sx=fun()+360*120
//13*sx=fun()-360*120
const double eps=1e-10;
const double eps2=1e-3;
int h0,m0,s0;
int t1,t2;
int ans[50];
int len;
double fun(int tmp)
{
double h,m,s;
s=tmp%(13*60)/13.0;
tmp/=13*60;
m=tmp%60;
h=tmp/60;
double ang0=(30*h0)+(m0/2.0)+(s0/120.0);
double ang1=(30*h)+(m/2.0)+(s/120.0);
double ang2=(6*m)+(s/10.0);
return ang1+ang2-2*ang0;
}
int bs(int l,int r,double x)
{
int mid;
while(r-l>1)
{
mid=(l+r)/2;
double cmp=fun(mid)-x;
if(cmp>=eps)
r=mid;
else
l=mid;
}
if(t1<=l&&l<=t2&&fabs(fun(l)-x)<eps2)
return l;
else if(t1<=r&&r<=t2&&fabs(fun(r)-x)<eps2)
return r;
return -1;
}
void solve(int i)
{
int l=i*46800,r=l+46800-1;
int tmp=bs(l,r,0.0);
if(tmp!=-1)
{
ans[len++]=tmp;
}
tmp=bs(l,r,360.0);
if(tmp!=-1)
{
ans[len++]=tmp;
}
tmp=bs(l,r,-360.0);
if(tmp!=-1)
{
ans[len++]=tmp;
}
}
int main()
{
//freopen("in","r",stdin);
//freopen("out","w",stdout);
int t,Case=0;
int h,m,s;
scanf("%d",&t);
while(t--)
{
len=0;
scanf("%d:%d:%d",&h0,&m0,&s0);
scanf("%d:%d:%d",&h,&m,&s);
t1=((h*60+m)*60+s)*13;
scanf("%d:%d:%d",&h,&m,&s);
t2=((h*60+m)*60+s)*13;
for(int i=0; i<12; i++)
{
solve(i);
}
sort(ans,ans+len);
len=unique(ans,ans+len)-ans;
printf("Case %d: %d\n",++Case,len);
int mod;
for(int i=0; i<len; i++)
{
mod=ans[i]%13;
ans[i]/=13;
s=ans[i]%60;
ans[i]/=60;
m=ans[i]%60;
h=ans[i]/60;
printf("%02d:%02d:%02d",h,m,s);
if(mod)
printf(" %d/13",mod);
puts("");
}
}
}
时间： 2024-10-10 01:02:02