http://codeforces.com/gym/101464/
过了5题
J： solved by ?
/*
ID: oodt
PROG:
LANG:C++
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<cassert>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxx=10005;
const int INF = 0x3f3f3f3f;
int n,m,k;
int a[maxx];
int vis[maxx];
int ans = 0,cnt = 0,pos = 0;
int l = 0,r = 0;
int main()
{
#ifdef LOCAL
// freopen("","r",stdin);
#endif
freopen("j.in","r",stdin);
while(~scanf("%d%d",&m,&n))
{
printf("%d\n",n/(m+1));
}
return 0;
}E： solved by ?
/*
ID: oodt
PROG:
LANG:C++
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<cassert>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxx=10005;
const int INF = 0x3f3f3f3f;
int n,m,k;
int a[maxx];
int vis[maxx];
int ans = 0,cnt = 0,pos = 0;
int l = 0,r = 0;
int main()
{
#ifdef LOCAL
// freopen("","r",stdin);
#endif
freopen("e.in","r",stdin);
int n;
int i;
double covermax;
int ans;
while(scanf("%d",&n)!=EOF)
{
ans=0;
double x;
for(i=1;i<=n;i++)
{
scanf("%lf",&x);
if(i==1){
covermax=x+1;
ans++;
continue;
}
else
{
if(x<covermax)continue;
else
{
ans++;
covermax=x+1;
}
}
}
printf("%d\n",ans);
}
return 0;
}H： solved by ?
/*
ID: oodt
PROG:
LANG:C++
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<cassert>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxx=1005;
const int INF = 0x3f3f3f3f;
int n,m,k;
int b[maxx][maxx];
int c[maxx][maxx];
struct pp{
ll h,l;
}a[maxx][maxx];
int vis[maxx];
int ans = 0,cnt = 0,pos = 0;
int l = 0,r = 0;
int main()
{
#ifdef LOCAL
// freopen("","r",stdin);
#endif
freopen("h.in","r",stdin);
while(~scanf("%d",&n))
{
memset(a,0,sizeof(a));
memset(c,0,sizeof(c));
rep(i,1,n+1)
rep(j,1,n+1)
scanf("%d",&b[i][j]);
rep(i,1,n+1){
rep(j,1,n+1){
a[i][j].h = a[i][j-1].h + b[i][j];
}
}
rep(j,1,n+1){
rep(i,1,n+1){
a[i][j].l = a[i-1][j].l + b[i][j];
}
}
rep(i,1,n+1){
rep(j,1,n+1){
if((a[i][j].l & 1) && (a[i][j].h&1))
{
c[i][j] = 0;
}
if((a[i][j].l & 1) == 0 && (a[i][j].h & 1) == 0){
if(c[i][j-1] && c[i-1][j]) c[i][j] = 0;
else c[i][j] = 1;
}
if((a[i][j].l & 1) == 0 && (a[i][j].h & 1) == 1){
c[i][j] = (c[i][j-1] + 1) % 2;
}
if((a[i][j].h & 1) == 0 && (a[i][j].l & 1) == 1){
c[i][j] = (c[i-1][j] + 1) % 2;
}
}
}
if(c[n][n] == 1) printf("W\n");
else printf("L\n");
}
return 0;
}A：高精度，用java写的，但是要求是读文件的操作啊，作为一个只会Biginteger的人，不会java的重定向，研究了将近2小时才研究出来, solved by lyy
import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) throws FileNotFoundException {
Scanner in=new Scanner(new File("A.IN"));
//Scanner in=new Scanner(System.in);
BigInteger a[]=new BigInteger[105];
int n,m;
while (in.hasNext())
{
m=in.nextInt();
n=in.nextInt();
for (int i=1;i<=m;i++)
{
a[i]=BigInteger.ONE;
}
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
{
BigInteger x=in.nextBigInteger();
a[j]=a[j].multiply(x);
}
}
BigInteger ans=a[1];
int cnt=1;
for (int i=1;i<=m;i++)
{
if (ans.max(a[i]).equals(a[i]))
{
ans=ans.max(a[i]);
cnt=i;
}
}
System.out.println(cnt);
}
}
}B：还是有高精度，而且还有表达式的处理，所以就偷个懒用python写，eval真的好用2333, solved by lyy
f=open('b.in','r')
data2=f.readlines()
w=''
for line in data2:
for s in line:
if s!=' ' and s!='\t' and s!='\r':
w=w+s
left=0
data=[]
z=''
for s in w:
if s=='\n':
if z!='' and z[-1]>='0'and z[-1]<='9':
data.append(z)
z='';
elif s=='(':
if left==0 and z!='':
data.append(z)
z=''
z=z+s
left+=1
elif s==')':
z=z+s
left-=1
if left==0 and z!='':
data.append(z)
z=''
else:
z=z+s
if (z!=''):
data.append(z)
for line in data:
#print(line)
s=line
s=s.replace('/','//')
l=len(s)
cnt=0;
for i in range(l):
if s[i]=='(':
cnt+=1
try:
r=0
for i in range(l):
if s[i]>='0' and s[i]<='9':
r+=1
else:
r=0
if r>90:
x=1//0
for step in range(cnt):
ss=''
l=len(s)
for i in range(l):
if s[i]=='(':
ss='('
elif s[i]==')':
ss=ss+')'
x=eval(ss)
if (x<0):
x=1//0
x=str(x)
if (len(x)>90):
x=1//0;
s=s.replace(ss,x)
#print(s)
break
else:
ss=ss+s[i]
print(s)
except:
print('Error')