到了与队友合作比拼的时刻了，这场可能将会是一场关乎icpc现场赛名额的比赛呢，虽然如今已经拿到了名额，但现在回忆起来，这场比赛确实还是意义非凡呢
作为这场参赛队伍中唯一一只拿到现场赛名额的队伍，然而，其实感觉这场很多题目都是可做的，还是缺乏配合吧。现在想想，最后的icpc现场赛的名额还是靠乌鲁木齐网赛rank1超越校金牌队的逆天表现而拼下来的，也算是一次难以忘怀的回忆了吧
1240 Problem D Godv与女朋友赛马
思路：典型的签到题，两数组排个序比较一下就ok了，从数模场强行赶往团队赛赛场，惊奇的发现队友已经把这题ac了，确实喜出望外啊
/*
Author Owen_Q
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
using namespace std;
const int maxn = 1e5+10;
int a[maxn],b[maxn];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int score = 0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
sort(a,a+n);
for(int i=0;i<n;i++)
{
scanf("%d",&b[i]);
}
sort(b,b+n);
for(int apos=0,bpos=0;apos<n&&bpos<n;apos++,bpos++)
{
while(a[apos]>=b[bpos])
{
bpos++;
if(bpos>=n)
{
break;
}
}
if(bpos<n)
{
score++;
}
}
if(score==0)
{
printf("Godv too strong\n");
}
else
{
printf("%d\n",score);
}
}
return 0;
}
1241 Problem E 余神的rp机
思路：经典dp
区间处理问题，那就先预处理区间最大值
之后按问题dp，对于m个核芯，n个时间段，dp由少一个核芯转移而来
/*
Author Owen_Q
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
using namespace std;
const int maxn = 510;
int s[maxn];
int inter[maxn][maxn];
int dp[maxn][maxn];
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&s[i]);
}
memset(inter,0,sizeof(inter));
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
inter[i][i] = s[i];
for(int j=i+1;j<=n;j++)
{
inter[i][j] = min(inter[i][j-1],s[j]);
}
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
dp[i][j] = dp[i-1][j];
for(int k=0;k<=j-1;k++)
{
dp[i][j] = max(dp[i][j],dp[i-1][k]+inter[k+1][j]*(j-k));
}
}
for(int j=2;j<=n;j++)
{
dp[i][j] = max(dp[i][j],dp[i][j-1]);
}
}
printf("%d\n",dp[m][n]);
/*for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
printf("%d ",dp[i][j]);
}
printf("\n");
}*/
}
}
return 0;
}
再加个小tip，通过本题第一次写了对拍的bat，瞬间感觉自己又学到了不少东西，来看看吧
#Author Owen_Q
@echo off
:again
rand.exe %random% > in.txt
standard.exe < in.txt > output.txt
余神的rp机.exe < in.txt > my.txt
fc /A output.txt my.txt
if not errorlevel 1 goto again
pause
goto again
然后加个随机数生成器
/*
Author Owen_Q
*/
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<sstream>
#include<ctime>
using namespace std;
#define random(a,b) ((a)+rand()%((b)-(a)+1))
stringstream ss;
int main(int argc, char *argv[])
{
int seed=time(NULL);
if(argc)//如果有参数
{
ss.clear();
ss<<argv[1];
ss>>seed;//把参数转换成整数赋值给seed
}
srand(seed);
//以上为随机数初始化，请勿修改
//random(a,b)生成[a,b]的随机整数
int t = random(1,5);
cout << t << endl;
while(t--)
{
int n,m;
n = random(1,10);
m = random(1,10);
while(m>n)
{
m = random(1,10);
}
cout << n << " " << m << endl;
for(int i=1;i<n;i++)
{
int s = random(0,500);
cout << s << " " ;
}
cout << random(0,500) << endl;
}
return 0;
}
完美
1243 Problem G CKJ老师爱数学
思路：这题当几何题做了好久，基本耽搁了整场，然而到头来才发现原来是个数论题，其实原来也想过优化枚举，但一旦思路跑偏了，就再也回不来了
简单的思路写代码前面了，直接看吧
/*
Author Owen_Q
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
/*
x^2 + y^2 = z^2
x^2 = z^2 - y^2
x^2 = (z-y) * (z+y)
令 d = gcd(z-y,z+y)
故 (z-y)/d 与 (z+y)/d 互质
则 z-y = d * u^2
z+y = d * v^2
因此
2z = d * (u^2 + v^2)
2y = d * (u^2 - v^2)
x = d * u * v
u v 互质
*/
using namespace std;
const long long maxn = 1e5+10;
vector <long long> factor;
void get_factor(long long x)
{
factor.clear();
long long i;
for(i=1;i*i<x;i++)
{
if(x%i==0)
{
factor.push_back(i);
factor.push_back(x/i);
}
}
if(i*i==x)
{
factor.push_back(i);
}
return ;
}
long long gcd(long long a,long long b)
{
return b == 0 ? a : gcd(b, a % b);
}
long long isSquare(long long x)
{
long long i = sqrt((double)(x+1e-9));
if(i*i == x)
{
return i;
}
else
{
return -1;
}
}
int main()
{
long long z;
int t;
scanf("%d",&t);
//freopen("in.txt","r",stdin);
//freopen("my.txt","w",stdout);
/*for(long long i=1;i<=2000;i++)
{
cout << i << endl;
}*/
while(t--)
{
scanf("%lld",&z);
long long sum = 0;
get_factor(2*z);
long long len = factor.size();
//cout << "*" << endl;
for(long long d=0;d<len;d++)
{
//cout << factor[d] << endl;
for(long long u=1;u*u<z/factor[d];u++)
{
//cout << "**" << endl;
long long v = isSquare(2*z/factor[d] - u*u);
if(v > 0&&gcd(u,v)==1)
{
sum++;
//cout << (v*v-u*u) * factor[d] / 2 << "**" << u*v*factor[d] << endl;
//cout << u <<"*"<< v <<"*"<< factor[d]<<endl;
}
}
}
sum *= 4;
sum += 4;
printf("%lld\n",sum);
}
return 0;
}
1239 Problem C Glory And Xor/Or
思路：这题其实是典型团队合作的成果，队友想到枚举所有数判断，然而n为1e5，完全可以加个小数量级的复杂度，果断想到按位判断，从高位开始，依次处理。
而异或问题其实就相当于分割问题，变相插板法的排列组合
最后就是要注意一下移位操作
/*
Author Owen_Q
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int last[maxn];
int now[maxn];
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
int n,k;
int re = 0;
scanf("%d%d",&n,&k);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
if(k==0)
{
printf("0\n");
continue;
}
memset(now,0,sizeof(now));
memset(last,0,sizeof(last));
for(int i=30;i>=0;i--)
{
///cout << endl << now[0] << " ";
now[0] = (a[0]&(1<<i))>>i;
for(int j=1;j<n;j++)
{
now[j] = now[j-1] ^ ((a[j]&(1<<i))>>i);
//cout << now[j] << "*" << (a[j]&(1<<i)) << "*" << i << " ";
}
if(now[n-1]==1)
{
re |= (1<<i);
//cout << "*";
}
else
{
int num = 0;
for(int j=0;j<n;j++)
{
now[j] |= last[j];
if(now[j]==0)
{
num++;
}
}
if(num>=k)
{
for(int j=0;j<n;j++)
{
last[j] = now[j];
}
//cout << "$";
}
else
{
re |= (1<<i);
//cout << "*";
}
}
/*for(int i=0;i<n;i++)
{
//cout << last[i] << " ";
}*/
}
printf("%d\n",re);
}
}
return 0;
}
icpc之旅奋斗启程
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