2020年蓝桥杯省赛 C++ B组
门牌制作
签到
#include
#include
using namespace std;
int n,m;
int get(int x)
{
int cnt = 0;
while(x)
{
if(x%10 == 2) cnt++;
x/=10;
}
return cnt;
}
int main()
{
n= 2020;
int ans = 0;
for(int i = 1; i <= n; i ++)
{
ans += get(i);
}
cout << ans << endl;
return 0;
} 既约分数
知识点：最大公约数gcd
#include
#include
using namespace std;
int n,m;
int gcd(int a,int b)
{
if(b == 0) return a;
return gcd(b,a%b);
}
int main()
{
n= 2020;
int ans = 0;
for(int i = 1; i <= n; i ++)
{
for(int j = 1; j <= n; j ++)
{
if(gcd(i,j) == 1) ans ++;
}
}
cout << ans << endl;
return 0;
} 蛇形填数
知识点：曼哈顿距离
#include
#include
#include
using namespace std;
int n,m;
int a[100][100];
int main()
{
int cnt = 1;
n = 80;
for(int i = 2; i <= n; i ++)
{
if(i%2)
{
for(int j = 1; j < i; j ++)
{
a[j][i-j] = cnt++;
}
}
else
{
for(int j = i - 1; j >= 1; j --)
{
a[j][i-j] = cnt++;
}
}
}
cout << a[20][20] << endl;
return 0;
} 七段码
知识点：连通块，二进制枚举
abcdefg分别对应二进制上每一位，共7位，0表示不存在，1表示存在；
建立各点的连接关系，对每一次枚举的数据查找连通块，如果连通块数量大于2，就不符合题意；
#include
#include
#include
using namespace std;
int n,m;
int e[100],ne[100],h[100],idx;
void add(int a,int b)
{
e[idx] = b,ne[idx] = h[a], h[a] = idx ++;
}
bool st[100];
void dfs(int x,int pos)
{
for(int i = h[pos]; i != -1 ; i = ne[i])
{
int j = e[i];
if(x&(1<<(j-1)) && !st[j])
{
// cout << pos << " " << j << " ";
st[j] = true;
dfs(x,j);
}
}
return;
}
int check(int x)
{
int cnt = 0;
// cout << x << " ";
memset(st,false,sizeof st);
for(int i = 0; i < 7; i ++)
{
if(x&(1< 跑步锻炼
知识点：闰年，模拟
#include
#include
#include
using namespace std;
int n,m;
int month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
// 0 1 2 3 4 5 6 7 8 9 10 11 12
int r_month[] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
int check(int x,int y)
{
if(x%4 == 0 || x%400 == 0)
{
return r_month[y];
}
return month[y];
}
int main()
{
int sun = 6;
int cnt = 0;
for(int i = 2000; i <= 2020; i ++)
{
for(int j = 1; j <= 12; j ++)
{
for(int k = 1; k <= check(i,j); k++)
{
if(sun == 8)
{
sun = 1;
}
if(k == 1 || sun == 1)
{
cnt ++;
}
cnt ++;
sun++;
if(i == 2020 && j == 10 && k == 1)
{
cout << cnt << endl;
return 0;
}
}
}
}
return 0;
} 回文日期
暴力枚举
#include
#include
#include
#include
#include
using namespace std;
int n,m;
int month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
// 0 1 2 3 4 5 6 7 8 9 10 11 12
int r_month[] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
vector q1,q2;
int check(int x,int y)
{
if((x%4 == 0 && x%100 != 0) || x%400 == 0)
{
return r_month[y];
}
return month[y];
}
bool pd1(int x)
{
int a[10];
int cnt = 1;
while(x)
{
a[cnt] = x % 10;
cnt++;
x /= 10;
}
for(int i = 1, j = 8; i <= j; i ++,j --)
{
if(a[i] != a[j])
{
return false;
}
}
return true;
}
bool pd2(int x)
{
int a[10];
int kind[10];
int cnt = 1;
for(int i = 0; i <= 9; i ++) kind[i] = 0;
while(x)
{
int t= x % 10;
kind[t] ++;
a[cnt] = t;
cnt++;
x /= 10;
}
int kinds = 0,last = -1;
for(int i = 0; i <= 9; i++)
{
if(kind[i])
{
kinds ++;
last = i;
}
}
if(kinds == 2)
{
string s;
for(int i = 1; i <= 8; i ++)
{
if(a[i] == a[1])
{
s += "A";
}
else
{
s += "B";
}
}
if(s == "ABABBABA")
{
return true;
}
}
return false;
}
void solve()
{
cin >> n;
int pos1 = upper_bound(q1.begin(),q1.end(),n)-q1.begin();
int pos2 = upper_bound(q2.begin(),q2.end(),n)-q2.begin();
cout << q1[pos1] << "\n" << q2[pos2] << endl;
return;
}
int main()
{
for(int i = 1; i <= 9999; i ++)
{
for(int j = 1; j <= 12; j ++)
{
for(int k = 1; k <= check(i,j); k++)
{
int cnt = i * 10000 + j *100 + k;
if(pd1(cnt))
{
q1.push_back(cnt);
}
if(pd2(cnt))
{
q2.push_back(cnt);
}
}
}
}
int t;
cin >> t;
while(t--)
{
solve();
}
return 0;
} 字串排序
蒟蒻还不会
成绩统计
#include
#include
#include
using namespace std;
int n,m;
int main()
{
double ji = 0, you = 0;
cin >> n;
for(int i = 1; i <= n; i ++)
{
int t;
cin >> t;
if(t >= 60) ji ++;
if(t >= 85) you ++;
}
double l = ji/(double)n;
double r = you/(double)n;
printf("%.0lf%%\n",round(l*100));
printf("%.0lf%%\n",round(r*100));
return 0;
} 子串分值和
知识点：
#include
#include
#define int long long
using namespace std;
int n;
int last[300];
string s;
signed main()
{
cin >> s;
n = s.size();
s = " " + s;
int ans = 0;
for(int i = 1; i <= n; i ++)
{
ans += (i-last[s[i]] ) * (n - i + 1);
last[s[i]] = i;
}
cout << ans << endl;
return 0;
} 平面切分
知识点：点与直线，数论
现有交点n条
与现有的直线相交时，新增的交点m个，答案贡献n+m+1
#include
#include
#include
#include
#include
#define int long long
using namespace std;
typedef pair PDD;
const int N = 2e5 + 10;
int n,m;
int a[N],b[N];
set s;
signed main()
{
cin >> n;
for(int i = 1; i <= n; i ++)
{
cin >> a[i] >> b[i];
s.insert({a[i],b[i]});
}
int ans = 2,cnt = 0;
for(auto i:s)
{
a[cnt] = i.first;
b[cnt] = i.second;
cnt ++;
}
for(int i = 1; i < s.size(); i ++)
{
double x = a[i];
double y = b[i];
set t;
for(int j = 0; j < i; j ++)
{
double l = a[j];
double r = b[j];
if(l == x)
continue;
double x1 = -(r - y)/(l - x);
double y1 = x1 * x + y;
t.insert({x1,y1});
}
ans += t.size() + 1;
}
cout << ans << endl;
return 0;
}