CodeForces - 1393
A - Rainbow Dash, Fluttershy and Chess Coloring
手动算几个
int t,n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
W(n/2+1);
}
}
B - Applejack and Storages
统计当前边长的个数，由于情况很少，全部枚举出来
int n,x,num[maxn],q;
char s[4];
int main()
{
int num2=0,num4=0,num6=0,num8=0;
scanf("%d",&n);
rep(i,1,n)
{
scanf("%d",&x);
num[x]++;
if (num[x]==8)num8++,num6--;
else if (num[x]==6)num6++,num4--;
else if (num[x]==4)num4++,num2--;
else if (num[x]==2)num2++;
}
scanf("%d",&q);
while(q--)
{
scanf("%s",s);
scanf("%d",&x);
if (s[0]=='+')
{
num[x]++;
if (num[x]==8)num8++,num6--;
else if (num[x]==6)num6++,num4--;
else if (num[x]==4)num4++,num2--;
else if (num[x]==2)num2++;
}
else if (s[0]=='-')
{
if (num[x]==8)num8--,num6++;
else if (num[x]==6)num6--,num4++;
else if (num[x]==4)num4--,num2++;
else if (num[x]==2)num2--;
num[x]--;
}
if (num8||num6>=2||num4>=2)printf("YES\n");
else if (num6&&num2)printf("YES\n");
else if (num6&&num4)printf("YES\n");
else if (num4&&num2>=2)printf("YES\n");
else printf("NO\n");
}
return 0;
}
C - Pinkie Pie Eats Patty-cakes
贪心
把数量最多的的第一个放在第一个位置,共有maxx-1个间隔
如果数量最多并不唯一，就放在结尾，例如123123123
int t,num[maxn],n,x;
int main()
{
scanf("%d",&t);
while(t--)
{
int maxx=0,cnt=0;
mem(num,0);
scanf("%d",&n);
rep(i,1,n)
{
scanf("%d",&x);
num[x]++;
maxx=max(maxx,num[x]);
}
rep(i,1,n)if (num[i]==maxx)cnt++;
n-=(cnt-1);
W((n-maxx)/(maxx-1));
}
}
D - Rarity and New Dress
dp[i][j]表示以mapp[i][j]为最低点的菱形个数
为什么dp关系是由dp[i-2][j],dp[i-1][j-1]和dp[i-1][j+1]控制的
而不是由dp[i-1][j],dp[i-1][j-1]和dp[i-1][j+1]控制的
因为每次菱形边长+1,会发现他的竖直长度+2
也就是说现在的[i,j]位置是由[i-2][j]位置推过来的
int dp[maxn][maxn],ans=0,n,m;
int main()
{
scanf("%d%d",&n,&m);
rep(i,0,n-1)scanf("%s",mapp[i]);
rep(i,0,n-1)
{
rep(j,0,m-1)
{
dp[i][j]=1;
if (i<2||j==0||j==m-1)dp[i][j]=1;
else if (mapp[i][j]==mapp[i-1][j]&&mapp[i][j]==mapp[i-1][j-1]&&mapp[i][j]==mapp[i-2][j]&&mapp[i][j]==mapp[i-1][j+1])
dp[i][j]+=min(min(dp[i-1][j-1],dp[i-2][j]),dp[i-1][j+1]);
ans+=dp[i][j];
}
}
W(ans);
return 0;
}
