The 2016 ACM-ICPC Asia Tehran Regional Contest 部分题解！

The 2016 ACM-ICPC Asia Tehran Regional Contest

  

A - Tax

 UVALive
- 7815

int n;
while(~scanf("%d",&n)&&n)
{
double ans=0;
if(n>1000000&&n<=5000000) ans+=n*0.1;
if(n>5000000) ans=n*0.2;
printf("%d\n",int(n-ans));
}

B - Key Maker

int n,m;
while(~scanf("%d%d",&m,&n))
{
if(n==0&&m==0) return 0;
for(int i=1; i<=m; i++) scanf("%d",&a[i]);
int ans=0;
for(int i=1; i<=n; i++)
{
int f=1,x;
for(int j=1; j<=m; j++)
{
scanf("%d",&x);
if(x>a[j]) f=0;
}
if(f) ans++;
}
pd(ans);
}

C - IOI 2017 Logo

 按每个log得分排序，得分相同按三分排序，按二分排序，按下标排序。

D - MicroRNA Ranking

int m,n;
int a[210]
,w

;
void init()
{
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++) scanf("%d",&a[i][j]);
cls(w,0);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
for(int k=j+1; k<=m; k++)
w[a[i][j]][a[i][k]]++;
}
for(int i=1; i<=m; i++)
for(int j=1; j<=m; j++)
{
if(w[i][j]>=(n+1)/2) w[i][j]=1;
else w[i][j]=0;
}
}
vector<int>ans;
void solve()
{
init();
int num=1,flag=0;
ans.clear();
ans.push_back(1);
for(int i=2; i<=m; i++)
{
int f1=-1,f2=0;
for(int j=0; j<num; j++)
if(!w[ans[j]][i])//可以插入这个位置
{
f1=j;
break;
}
if(f1==-1) f1=num;
for(int j=f1; j<num&&!f2; j++) if(w[ans[j]][i]&&!w[i][ans[j]]) f2=1;
if(f2) flag=1;
ans.insert(ans.begin()+f1,i);
num++;
}
if(flag) puts("No solution");
else for(int i=0; i<m; i++) printf("%d%c",ans[i],i==m-1?'\n':' ');
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
if(!m&&!n) return 0;
solve();
}
return 0;
}

//5 3
//3 2 4 1 5
//4 1 5 2 3
//2 4 5 1 3
//
//5 2
//5 4 3 2 1
//1 2 3 4 5
//
//4 3
//1 4 2 3
//4 2 3 1
//3 1 2 4
//
//4 4
//1 2 3 4
//1 2 3 4
//1 4 3 2
//1 4 3 2
//
//4 4
//1 2 3 4
//1 2 3 4
//1 3 4 2
//1 3 4 2
//
//4 4
//1 2 3 4
//1 2 3 4
//1 4 3 2
//1 4 3 2
G - Elections

int n;
while(~scanf("%d",&n)&&n)
{
memset(dp,0,sizeof(dp));
dp[0][0]=1.0;
int sum=0;
for(int i=1; i<=n; i++)
{
scanf("%lf%d",&a[i],&b[i]);
sum+=b[i];
}
for(int i=1; i<=n; i++)
for(int j=0; j<=sum; j++)
if(dp[i-1][j])
{
dp[i][j]+=(1.0-a[i])*dp[i-1][j];
dp[i][j+b[i]]+=a[i]*dp[i-1][j];
}
double ans=0.0;
for(int i=(sum+1)/2; i<=sum; i++)
ans+=dp
[i];
printf("%.4f\n",ans);
}