感想：
今天三个人的状态比昨天计院校赛的状态要好很多，然而三个人都慢热体质导致签到题wa了很多发。最后虽然跟大家题数一样(6题)，然而输在罚时。
只能说，水题还是刷得少，看到签到都没灵感实在不应该。
题目链接：http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=391
A：简单贪心，按高度sort一下就好了，这里用优先队列处理
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional> using namespace std; int M[], F[]; int main(void)
{
int T;
while (scanf("%d", &T) != EOF)
{
while (T--)
{
int n, m;
priority_queue <int, vector<int>, greater <int>> fu, fd, mu, md;
scanf("%d %d", &n, &m);
for (int i = ; i < n; i++)
{
scanf("%d", &M[i]);
}
for (int i = ; i < m; i++)
{
scanf("%d", &F[i]);
}
for (int i = ; i < n; i++)
{
int flag;
scanf("%d", &flag);
if (flag)
{
mu.push(M[i]);
}
else
{
md.push(M[i]);
}
}
for (int i = ; i < m; i++)
{
int flag;
scanf("%d", &flag);
if (flag)
{
fu.push(F[i]);
}
else
{
fd.push(F[i]);
}
}
int ans = ;
// fu <-> md
while (!fu.empty() && !md.empty())
{
if (fu.top() < md.top())
{
// printf("pop: %d %d\n", fu.top(), md.top());
fu.pop();
md.pop();
ans++;
}
else
{
// printf("pop: %d\n", md.top());
md.pop();
}
}
// fd <-> mu
while (!mu.empty() && !fd.empty())
{
if (mu.top() < fd.top())
{
// printf("pop: %d %d\n", mu.top(), fd.top());
mu.pop();
fd.pop();
ans++;
}
else
{
// printf("pop: %d\n", fd.top());
fd.pop();
}
}
printf("%d\n", ans);
}
}
return ;
}
B：找规律，显然是不停把n除二加起来，高精就用java
import java.util.*;
import java.lang.*;
import java.math.BigInteger; public class Main { public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
while (cin.hasNextInt())
{
int t=cin.nextInt();
while (t-->0){
BigInteger ans=BigInteger.ZERO;
BigInteger x=cin.nextBigInteger();
while (!x.equals(BigInteger.ONE)){
x=x.divide(BigInteger.valueOf(2));
ans=ans.add(x);
}
System.out.println(ans);
}
}
}
}
C：模拟
#include <cstdio> char t[][];
bool vis[][]; const int p3[] = {, , , , , }; const int movement[][] =
{
{, }, // DOWN
{, }, // RIGHT
{-, }, // UP
{, -} // LEFT
}; void init(int n, int m)
{
for (int i = ; i < n; i++)
{
for (int j = ; j < m; j++)
{
vis[i][j] = false;
}
}
} int solve(void)
{
int ans = ;
int n, m;
int a, b;
long long int k;
scanf("%d %d", &n, &m);
scanf("%d %d %lld", &a, &b, &k);
a--;
b--;
char cmd[];
scanf(" %s", cmd);
init(n, m);
for (int i = ; i < n; i++)
{
scanf(" %s", t[i]);
for (int j = ; j < m; j++)
{
t[i][j] -= '';
}
}
while (k--)
{
int x = p3[] * t[a][b]
+ p3[] * t[a - ][b]
+ p3[] * t[a + ][b]
+ p3[] * t[a][b - ]
+ p3[] * t[a][b + ];
if (vis[a][b])
{
return ans;
}
vis[a][b] = true;
if (cmd[x] == 'D')
{
int na = a + movement[][], nb = b + movement[][];
if (t[na][nb] == ) return ans;
else a = na, b = nb;
}
else if (cmd[x] == 'R')
{
int na = a + movement[][], nb = b + movement[][];
if (t[na][nb] == ) return ans;
else a = na, b = nb;
}
else if (cmd[x] == 'U')
{
int na = a + movement[][], nb = b + movement[][];
if (t[na][nb] == ) return ans;
else a = na, b = nb;
}
else if (cmd[x] == 'L')
{
int na = a + movement[][], nb = b + movement[][];
if (t[na][nb] == ) return ans;
else a = na, b = nb;
}
else if (cmd[x] == 'P')
{
if (t[a][b] == )
{
t[a][b] = ;
ans++;
init(n, m);
}
}
else if (cmd[x] == 'I')
{
return ans;
}
}
return ans;
} int main(void)
{
int T;
while (scanf("%d", &T) != EOF)
{
while (T--)
{
printf("%d\n", solve());
}
}
return ;
}
D：上一题的人工智能版，要你构造特定程序捡垃圾。方法是走回字形，比如我们选定顺时针方向走，那么当我们走到左边靠墙位置的时候，如果右手边有垃圾，那么我们往右走一格再继续往上走。如果走到地图中间位置(四周没垃圾)就往上走。如果机器人走了连续相同方向n次就让机器人“抖动”一下(属实人工智能调参)。然而cy他们队就是A了(神仙啊
E：从右往左扫一次就好了，队友没开LL导致wa一发要批评
#include <cstdio> long long int a[], b[]; int main(void)
{
int T;
while (scanf("%d", &T) != EOF)
{
while (T--)
{
int n;
scanf("%d", &n);
for (int i = ; i < n; i++)
{
scanf("%lld", &a[i]);
}
for (int i = ; i < n; i++)
{
scanf("%lld", &b[i]);
}
bool flag = true;
for (int i = n - ; i >= ; i--)
{
if (b[i] >= a[i])
{
if (i)
{
b[i - ] += b[i] - a[i];
}
}
else
{
flag = false;
break;
}
}
printf(flag ? "Yes\n" : "No\n");
}
}
return ;
}
F：神仙题
G：非常简单的贪心
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional> using namespace std; int M[], F[]; int main(void)
{
int T;
while (scanf("%d", &T) != EOF)
{
while (T--)
{
int n, k;
scanf("%d %d", &n, &k);
priority_queue <long long int> pos, neg;
for (int i = ; i < n; i++)
{
int tmp;
scanf("%d", &tmp);
if (tmp > )
{
pos.push(tmp);
}
else
{
neg.push(-tmp);
}
}
long long int ans = , maxn = ;
while (!pos.empty())
{
ans += * pos.top();
maxn = max(pos.top(), maxn);
int cnt = k - ;
pos.pop();
while (cnt-- && !pos.empty())
{
pos.pop();
}
}
while (!neg.empty())
{
ans += * neg.top();
maxn = max(neg.top(), maxn);
int cnt = k - ;
neg.pop();
while (cnt-- && !neg.empty())
{
neg.pop();
}
}
printf("%lld\n", ans - maxn);
}
}
return ;
}
H：救公主，边双相关的题目(然而队友最后没撸出来)
I：要求逆元的看不懂的题目
J：签到(wa了7次，三个人都没睡醒。这里我马上想到了可以输出n×2和n×3，然而忘记特判n==1的情况……)
#include<iostream>
#include<cstdio>
#include<cstdlib> using namespace std; typedef long long ll; ll read()
{
ll x = ; char c = getchar(); ll flag = ;
while (c < '' || c > '')
{
if (c == '-')
{
flag = -;
}
c = getchar();
}
while (c >= '' && c <= '')x = x * 10ll + c - '', c = getchar();
return x;
} int main()
{
ll T;
while (scanf("%lld", &T) == )
{
while (T--)
{
ll x;
x = read();
if (x % == )
{
printf("4 %lld\n", + x);
}
else
{
printf("15 %lld\n", + x);
}
}
} return ;
}