2018多校联合训练4
http://acm.hdu.edu.cn/userloginex.php?cid=805
过了6题，第一次进入前100名
1012：直接从1走到n
solved by wyq
#include
#include
int a[100005];
int reabs(int x)
{
if(x<0)return -x;
else return x;
}
int main()
{
int T,i,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
printf("%d\n",(int)(sqrt(reabs(a[1]-a[n]))));
}
return 0;
}
1004：和出题人心有灵犀2333
solved by lyy
#include
using namespace std;
#define ll long long
struct st
{
int a,b;
}p[105];
int t,n,m;
bool cmp(st a,st b)
{
return a.b
1011：
solved by wyq
#include
#include
#include
char s[1005];
bool isnum(int x,int y)
{
if(x>y)return false;
//printf("%d %d\n",x,y);
if(x!=y && s[x]=='0')return false;
int i;
for(i=x;i<=y;i++)
if(s[i]=='+' || s[i]=='*')return false;
for(i=x;i<=y;i++)
{
if(s[i]=='?')s[i]='1';
}
return true;
}
int main()
{
int T,i,n;
scanf("%d",&T);
while(T--)
{
scanf("%s",s);
int lens=strlen(s);
if(s[0]=='0' && s[1]=='?')s[1]='+';
for(i=2;i<=lens;i++)
if((s[i-2]=='+' || s[i-2]=='*')&& s[i-1]=='0' && s[i]=='?')s[i]='+';
int lastfuhao=lens;
bool ans=true;
for(i=lens-1;i>=0;i--)
{
if(s[i]=='+' || s[i]=='*')
{
ans=isnum(i+1,lastfuhao-1);
lastfuhao=i;
if(!ans)break;
}
}
if(ans)ans=isnum(0,lastfuhao-1);
if(!ans)printf("IMPOSSIBLE\n");
else printf("%s\n",s);
}
return 0;
}
1002：搞不懂为什么是莫队算法，不过和莫队算法比较相似,按照
分块，首先求出所有200的倍数行，然后利用
递推
solved by lyy
#include
using namespace std;
#define ll long long
const ll mod=1000000007;
ll f[100005];
ll ff[100005];
ll c[205][100005];
int t,n,m;
long long inv(long long a,long long m)
{
if(a == 1)return 1;
return inv(m%a,m)*(m-m/a)%m;
}
int main()
{
f[0]=1;
for (int i=1;i<=100000;i++)
{
f[i]=f[i-1]*i%mod;
}
for (int i=0;i<=100000;i++)
{
ff[i]=inv(f[i],mod);
}
c[0][0]=0;
for (int i=1;i<=200;i++)
{
ll x=1;
c[i][0]=1;
for (int j=1;j<=500*i;j++)
{
x=(x+f[i*500]*ff[j]%mod*ff[500*i-j]%mod)%mod;
c[i][j]=x;
}
}
scanf("%d",&t);
while (t--)
{
scanf("%d%d",&n,&m);
if (m<=600)
{
ll x=1;
for (int i=1;i<=m;i++)
{
x=(x+f[n]*ff[i]%mod*ff[n-i]%mod)%mod;
}
printf("%lld\n",x);
}
else if (n==m)
{
ll ans=1;
for (int i=1;i<=n;i++)
{
ans=ans*2%mod;
}
printf("%lld\n",ans);
}
else
{
vector v;
ll ans=0;
while (n%500!=0)
{
v.push_back(f[n-1]*ff[m]%mod*ff[n-1-m]%mod);
m--;
n--;
}
reverse(v.begin(),v.end());
ans=c[n/500][m];
for (int i=0;i
1005：行和列都是2L的周期，第i行第j 列的数是：
那么很容易分析出i和j都是2L的周期，更确切的是当n位奇数时周期是L，n为偶数时周期为2L（一开始把周期当L，结果Wa了），
然后分成4块求即可
solved by lyy
#include
using namespace std;
#define ll long long
int t;
int l,q;
int a[22];
int p[22][22];
int r[22];
int row[22][22];
int col[22][22];
ll s[22];
int main()
{
scanf("%d",&t);
while (t--)
{
memset(p,0,sizeof(p));
memset(r,0,sizeof(r));
memset(s,0,sizeof(s));
memset(row,0,sizeof(row));
memset(col,0,sizeof(col));
scanf("%d",&l);
int k=l*2;
for (int i=0;i
1010：愉快的暴力题，先对行暴力，时间复杂度 O(4*4^5*16)，每行只有两种可行的情况，再判断列并取最小值即可
solved by lyy
#include
using namespace std;
#define ll long long
#define per(i,j,k) for (int i=j;i go(int r)
{
vector ans;
st x;
for (int i=0;i<4;i++)
{
for (int j=0;j<16;j++)
{
x.data[i][j]=a[i+r*4][j];
}
}
per(x1,0,4)
{
per(x2,0,4)
{
per(x3,0,4)
{
per(x4,0,4)
{
int can=1;
for (int i=0;i<4;i++)
{
init();
for (int j=0;j<16;j++)
{
if (vis[x.data[i][j]]==1) can=0;
else vis[x.data[i][j]]=1;
}
}
if (can==1)
{
//cout< v1=go(0);
vector v2=go(1);
vector v3=go(2);
vector v4=go(3);
int ans=10000;
for (auto x1:v1)
{
for (auto x2:v2)
{
for (auto x3:v3)
{
for (auto x4:v4)
{
int can=1;
for (int i=0;i<16;i++)
{
init();
for (int j=0;j<4;j++)
{
if (vis[x1.data[j][i]]==1) can=0;
else vis[x1.data[j][i]]=1;
}
for (int j=0;j<4;j++)
{
if (vis[x2.data[j][i]]==1) can=0;
else vis[x2.data[j][i]]=1;
}
for (int j=0;j<4;j++)
{
if (vis[x3.data[j][i]]==1) can=0;
else vis[x3.data[j][i]]=1;
}
for (int j=0;j<4;j++)
{
if (vis[x4.data[j][i]]==1) can=0;
else vis[x4.data[j][i]]=1;
}
if (can==1)
{
ans=min(ans,x1.num+x2.num+x3.num+x4.num);
}
}
}
}
}
}
printf("%d\n",ans);
}
return 0;
}