Preamble:
Happy winter holiday week-long training camp is finally over, nice to meet so many chiefs and seniors, sister school. It thanked ljw seniors and Chen Yu teacher training for our winter vacation made.
Simple comment about today's examination papers, just before the noon game, jlw seniors said the title is water, it is water, can you believe it? ? Anyway, I believe, in fact, is really very water (I'm too weak), ah ah ah, there are some questions giant pit ...... I did not talk much, first put about some cases of tonight's game now.
Information：
Ended up:
Rank List：
Black can, Jiang Shi-chao Black, light fruit, puzzles Bear, Old Wei, Sun Fei (fu) Yue really good strong ah! !
topic:
problem A: 28 factor (NEFU 2101)
The core idea: violence, first determine the maximum of a few 7 (the less he will be the smaller number of bits), and then dwell on, to see which conditions are met, I was too dishes, this is my last question of the AC, to think complicated!
#include <bits/stdc++.h>
using namespace std;
int main()
{
//ios::sync_with_stdio(false);(玄学如果不注释会re)
int n;
while (~scanf("%d", &n))
{
int sum2 = 0;
int flag = 0, ans1, ans2; //4,7
sum2 = n / 7; //7的最大数量
for (int i = sum2; i >= 0; i--)
{
int tmp = n;
if ((tmp - i * 7) % 4 == 0)
{
flag = 1;
ans1 = (tmp - i * 7) / 4; //标记4的个数
ans2 = i; //标记7的个数
break;
}
}
if (!flag)
cout << "xinganheixiong" << endl;
else
{
for (int i = 1; i <= ans1; i++)
cout << "4";
for (int i = 1; i <= ans2; i++)
cout << "7";
cout << endl;
}
}
return 0;
}problem B: Chen bonuses (NEFU 2077)
The core idea: simple structure sorted.
#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 5;
struct student
{
int xh, cy, eng, math;
int sum;
int flag;
} a[M];
bool cmp(student a, student b)
{
if (a.sum != b.sum)
return a.sum > b.sum;
else
{
if (a.eng != b.eng)
return a.eng > b.eng;
else
{
return a.flag > b.flag;
}
}
}
int main()
{
ios::sync_with_stdio(false);
int n;
while (cin >> n)
{
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++)
{
cin >> a[i].xh >> a[i].cy >> a[i].eng >> a[i].math;
a[i].flag = i;
a[i].sum += a[i].math + a[i].cy;
}
sort(a + 1, a + 1 + n, cmp);
int tmp;
if (n < 4)
tmp = n;
else
tmp = 4;
for (int i = 1; i <= tmp; i++)
cout << a[i].xh << " " << a[i].sum << endl;
}
return 0;
}problem C: Bob the cake (NEFU 2096)
The core idea: This question is similar to fractional integration money, simple questions, to be able to see through it positive or negative situation is the same, and soon will be able to do it;
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
int a, b, x;
int ans;
cin >> t;
while (t--)
{
cin >> a >> b;
x = fabs(a - b);
ans = 0;
while (x >= 5)
{
ans += x / 5;
x -= x / 5 * 5;
}
while (x >= 2)
{
ans += x / 2;
x -= x / 2 * 2;
}
while (x >= 1)
{
ans += x / 1;
x -= x / 1;
}
cout << ans << endl;
}
return 0;
}problem D: magical thing happened (NEFU 2090)
The core idea: use the stack, the stack due to less exercise, I direct a string to write.
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
char a[105], b[105];
while (cin >> a)
{
int sum = 0, L;
L = strlen(a);
for (int i = 0; i < L; i++)
{
b[sum] = a[i];
if (b[sum] == 'o' && b[sum - 1] == 'o') //变化
{
b[sum - 1] = 'O'; //移位
sum--; //两个变成一个
}
if (b[sum] == 'O' && b[sum - 1] == 'O') //删除
sum -= 2; //一次是删除两个
sum++;//继续进行
}
for (int i = 0; i < sum; i++)
cout << b[i];
cout << endl;
}
return 0;
}problem E: jwMM selected hotel (NEFU 2083)
The core idea: not violent, time out, the core ideas, simulation, transfer state; concern at this hotel if you can drink milk tea , if the drink, is updated to the position of number of the same color as the hotel so , ans + = current of the same color hotel the number of (other than this hotel); if you can not drink it, ans + = number of hotels the same color (the same color at this time the number of hotels is sure to meet the current drink tea in front of the hotel).
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int M = 2e5 + 5;
LL ans = 0;
int color[M] = {0};
int num[55] = {0}; //同颜色酒店的数量
int main()
{
ios::sync_with_stdio(false);
int n, m, pp, pos = 0; //pos是标记能喝奶茶位置的最大值
cin >> n >> m >> pp; //酒店个数，颜色数目，最低消费
for (int i = 1; i <= n; i++) //暂时要住的
{
int col, ppi;
cin >> col >> ppi;
color[i] = col; //储存第i个位置是什么颜色
if (ppi <= pp) //i酒店能奶茶，加上之前的所以同色的酒店就ok
{
for (int j = pos + 1; j <= i; j++) //更新每种颜色酒店的数量
num[color[j]]++;
pos = i; //更新位置
ans += num[col] - 1;
}
else //i酒店不能喝奶茶，就加之前同色酒店数量就行
ans += num[col];
}
cout << ans << endl;
return 0;
}problem F: jwMM archery game (NEFU 2078)
The core idea: gcd use (gcd is to represent a common factor), the pit may have xy is less than or equal to 0, but for this is my blood of a QAQ ~~ pit.
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
int x1, y1, x2, y2;
while (cin >> x1 >> y1)
{
int num[500] = {0}, z = 0;
cin >> x2 >> y2;
int tmp1, tmp2;
tmp1 = __gcd(x1, y1);
tmp2 = x2 - y2;
if (tmp2 <= 0) //这个判断十分重要，题干说是正倍数
cout << "N" << endl;
else
{
if (__gcd(tmp1, tmp2) == 1) //如果是1的话他就没有除1以外的公因子
cout << "N" << endl;
else
cout << "Y" << endl;
}
}
return 0;
}problem G: Dan play (NEFU 2082)
核心思路：二进制枚举，当时没有看出来，同时也要能清楚知道题干的叙述。
#include <bits/stdc++.h>
using namespace std;
struct lam
{
int kg[15];
} a[15];
int main()
{
ios::sync_with_stdio(false);
int n, m, tmp, tmp2;
while (cin >> n >> m) //n是开关数量,m是灯泡数量
{
long long ans = 0;
int pi[15] = {0};
memset(a, 0, sizeof(a));
for (int i = 1; i <= m; i++) //输入灯泡及其开关编号
{
cin >> tmp;
for (int j = 1; j <= tmp; j++)
{
cin >> tmp2;
a[i].kg[tmp2] = 1; //代表的是第ki编号的灯泡是否有相应的开关，即开关标记
}
}
for (int i = 1; i <= m; i++) //输入灯泡亮的程度；
cin >> pi[i];
for (int i = 0; i < (1 << n); i++) //枚举n（开关数量）
{
int dp[15] = {0};
for (int j = 0; j < n; j++)
{
if (i & (1 << j)) //枚举开关开闭的状态
dp[j + 1] = 1;
else
dp[j + 1] = 0;
}
int z, judge = 0;
for (int k = 1; k <= m; k++) //判断每一个是否成立
{
int sum = 0;
for (z = 1; z <= n; z++)
if (a[k].kg[z] && dp[z]) //ki编号的灯是否有相应的开关
sum++;
if (sum % 2 != pi[k]) //亮度的判断
{
judge = 1;
break;
}
}
if (!judge)
ans++;
}
cout << ans << endl;
}
return 0;
}problem H：分糖果(NEFU 2100)
核心思想：二分查找。你的ans一定会小于m值，目的是找到两个数的取余后的加和最大。
- 利用取余的性质
(a+b)%m=(a%m+b%m)%m，对输入进行处理。
- 从i=1，一直查到i=n，最理想中的返回值是m-1，所以对其当作切入点，利用lower_bound（）查找大于等于m-a【i】的坐标。
- 关键：细节处理，设查找的数是tmp，lower_bound（）的防越界处理，当a【i】中的所有数都小于tmp的时候，他的返回值是1+n，是会越界的，要进行后越界处理；当a【i】中的所有数都小于tmp的时候，他的返回值是1，但是你要找的是小于最近一个tmp，要是减去1也是会出界，要做前出界处理。
- 最后一步，如果没有取到x为m-a[i]-1的话就不一定是最优解 所以和最后一个数相加判断。
#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 5;
int a[M], m;
int main()
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--)
{
int n;
memset(a, 0, sizeof(a));
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
a[i] %= m;
}
sort(a + 1, a + 1 + n);
int tmp1, tmp2, ans = 0;
for (int i = 1; i <= n; i++)
{
tmp1 = m - a[i];
tmp2 = lower_bound(a + 1, a + 1 + n, tmp1) - a;
if (tmp2 - 1 > 0 && tmp2 != 1 + n) //双倍防越界
{
if (tmp2 - 1 != i)
ans = max(ans, a[tmp2 - 1] + a[i]);
else if (tmp2 - 2 > 0) //如果他俩是同一个位置，就要向前移一位，不能加本身的
ans = max(ans, a[tmp2 - 2] + a[i]);
}
if (i != n) //防止加不到最大值的时候
ans = max((a[i] + a[n]) % m, ans);
}
cout << ans << endl;
}
return 0;
}
problem I：抹发胶(NEFU 2088)
核心思路：暴力。
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
int t;
while (cin >> t)
{
int n, a[105];
while (t--)
{
int minn[105] = {0};
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
for (int j = 1; j < i; j++) //在输入的同时判断前一名是否小于后一名
{
if (a[i] > a[j])
minn[i]++;
}
}
for (int i = 1; i <= n; i++)
{
cout << minn[i];
if (i == n)
cout << endl;
else
{
cout << " ";
}
}
}
}
return 0;
}problem J：天哥的难题(NEFU 2085)
核心思路：数学证明题，后续证明。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
LL fastpow(LL a, LL b)
{
int res = 1;
while (b)
{
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int main()
{
ios::sync_with_stdio(false);
int m;
while (cin >> m)
{
cout << fastpow(3, m) << endl;
}
return 0;
}problem K：煊哥的数字游戏(NEFU 2084)
核心思路：防AK的题，巨无敌难的数学推导。异或是不进位的加法。
- sum=a1+a2+...+am，num=a1⊕a2⊕...⊕am。这样就是解sum+b=num⊕2*b。
- 若两边相等，则它的二进制每一位都要相等，模拟等式左边+b，等式右边⊕2*b的过程，从第一位逐个比较，直到sum=num。
- 当sum=num的时候，等式左边=sum+b=sum'，要解b，只需要sum'-sum即可。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int a[100005];
int main()
{
ios::sync_with_stdio(false);
int m;
LL sum = 0, num = 0, tmp, i = 0;
cin >> m;
for (int i = 1; i <= m; i++)
{
cin >> a[i];
sum += a[i];
num ^= a[i];
}
LL ans = sum; //记录原来的sum
num *= 2;
while (sum != num)
{
i++;
tmp = (LL)1 << i; //比较每一位是否相同
if (sum % tmp != num % tmp)
{
sum += tmp >> 1;
num ^= tmp;
}
}
cout << sum - ans << endl;
return 0;
}problem L：吃辣条
核心思路：快排+暴力。
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
int n;
while (cin >> n)
{
int a[105];
for (int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + 1 + n);
cout << a[n - 1] << " " << a[2] << endl;
}
return 0;
}
