I am a slow walker, but I never walk backwards!
比賽連結:http://acm.bnu.edu.cn/v3/contest_show.php?cid=5727
手速不夠。。。簡直了啊
A:大神們說打表找規律,然後就可以得到了。
<span style="font-size:18px;">#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <iomanip> #include <stdio.h> #include <string> #include <queue> #include <cmath> #include <stack> #include <time.h> #include <map> #include <set> #define eps 1e-8 #define M 1000100 #define LL long long //#define LL long long #define INF 0x3f3f3f #define PI 3.1415926535898 #define read() freopen("data1.in", "r", stdin) #define write() freopen("data1.out", "w", stdout); #define mod 1000000007 const int maxn = 1000010; char str[maxn]; using namespace std; int dp[maxn]; struct node { int a, b; }f[maxn]; int main() { int T; scanf("%d",&T); while(T--) { int n, k; cin >>n>>k; if(n%2) { if((k-1)%4 < 2) puts("B"); else puts("A"); } else { if(k%2) puts("F"); else if(k%4 == 0) puts("A"); else puts("B"); } } }</span>
B:大水題,(n-1)/n,分數要化簡到最簡式。
C:比賽的時候猜了一下沒敢交。。。後來才發現是對的啊。。。<span style="font-size:18px;">#include <stdio.h> #include <string> #include <queue> #include <cmath> #include <stack> #include <time.h> #include <map> #include <set> #define eps 1e-8 #define M 1000100 #define LL long long //#define LL long long #define INF 0x3f3f3f #define PI 3.1415926535898 #define read() freopen("data1.in", "r", stdin) #define write() freopen("data1.out", "w", stdout); #define mod 1000000007 const int maxn = 210; using namespace std; int main() { int T; scanf("%d",&T); while(T--) { LL n; scanf("%lld",&n); LL m = (n-1LL); LL t = __gcd(n, m); cout<<m/t<<"/"<<n/t<<endl; } }</span>
什麼都不加的話先手必勝,加了的話會構成環,如果是奇數環的話從第一個開始取先手必勝,偶數的環的話,先把環拆開,然後先手再取最大的可以必勝。總之,先手必勝。
<span style="font-size:18px;">#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <iomanip> #include <stdio.h> #include <string> #include <queue> #include <cmath> #include <stack> #include <time.h> #include <map> #include <set> #define eps 1e-8 #define M 1000100 #define LL long long //#define LL long long #define INF 0x3f3f3f #define PI 3.1415926535898 #define read() freopen("data1.in", "r", stdin) #define write() freopen("data1.out", "w", stdout); const int maxn = 210; using namespace std; int main() { int T; cin >>T; while(T--) { int n, m; int x, y; cin >>n>>m; for(int i = 0; i < m; i++) scanf("%d %d",&x, &y); puts("Bill will lose HAHA"); } return 0; } </span>
D:矩陣加速遞推,水題做的太慢到這裡沒時間了啊。。sad
f(x)代表發育期;
g(x)代表更年期;
h(x)代表天數;
f(x) = y*g(x-1)+h(x-1)*p;
g(x) = x*f(x-1)+z*g(x-1);
h(x) = h(x-1)+1;
從:f(x-1) g(x-1) h(x-1) 1 到 f(x) g(x) h(x) 1
關係矩陣:
0 y p 0
x z 0 0
0 0 1 1
0 0 0 1
<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <time.h>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define read() freopen("data1.in", "r", stdin)
#define write() freopen("data1.out", "w", stdout);
const int maxn = 210;
using namespace std;
struct matrix
{
LL f[10][10];
};
matrix mul(matrix a, matrix b, int n)
{
matrix c;
memset(c.f, 0, sizeof(c.f));
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
for(int k = 0; k < n; k++) c.f[i][j] += a.f[i][k]*b.f[k][j];
c.f[i][j];
}
}
return c;
}
matrix pow_mod(matrix a, LL b, int n)
{
matrix s;
memset(s.f, 0 , sizeof(s.f));
for(int i = 0; i < n; i++) s.f[i][i] = 1LL;
while(b)
{
if(b&1) s = mul(s, a, n);
a = mul(a, a, n);
b >>= 1;
}
return s;
}
matrix Add(matrix a,matrix b, int n)
{
matrix c;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
c.f[i][j] = a.f[i][j]+b.f[i][j];
c.f[i][j];
}
}
return c;
}
int main()
{
int T;
cin >>T;
while(T--)
{
LL x, y, z, p, n;
cin >>x>>y>>z>>p>>n;
matrix c;
memset(c.f, 0 ,sizeof(c.f));
c.f[0][0] = 0;
c.f[0][1] = y;
c.f[0][2] = p;
c.f[0][3] = 0;
c.f[1][0] = x;
c.f[1][1] = z;
c.f[1][2] = 0;
c.f[1][3] = 0;
c.f[2][0] = 0;
c.f[2][1] = 0;
c.f[2][2] = 1;
c.f[2][3] = 1;
c.f[3][0] = 0;
c.f[3][1] = 0;
c.f[3][2] = 0;
c.f[3][3] = 1;
matrix d = pow_mod(c, n, 4);
LL sum = 0LL;
sum += d.f[0][3];
sum += d.f[1][3];
printf(">>%lld\n",sum);
}
return 0;
}
</span>
E:就是一個活動排序,大水題。
<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <time.h>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define read() freopen("data1.in", "r", stdin)
#define write() freopen("data1.out", "w", stdout);
#define mod 1000000007
const int maxn = 210;
using namespace std;
typedef struct sss
{
int start_time;
int end_time;
} s;
bool operator <(s a, s b)
{
return a.end_time < b.end_time;
}
int main()
{
int T;
cin >>T;
int n;
while(T--)
{
cin>>n;
int timestart = 0;
vector<s> a;
while(n--)
{
s temp;
cin>>temp.start_time>>temp.end_time;
a.push_back(temp);
}
sort(a.begin(),a.end(),less<s>());
int count = 0;
for(vector<s>::iterator ite = a.begin(); ite != a.end(); ++ite)
{
if((*ite).start_time > timestart)
{
count++;
timestart = (*ite).end_time;
}
}
cout<<count<<endl;
//a.clear();
}
}</span>
F:讀懂題之後就是一個標準的01揹包了啊。
PS:程式碼寫了1234B好巧啊。
<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <time.h>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define read() freopen("data1.in", "r", stdin)
#define write() freopen("data1.out", "w", stdout);
#define mod 1000000007
const int maxn = 1000010;
char str[maxn];
using namespace std;
int dp[maxn];
struct node
{
int a, b;
}f[maxn];
int main()
{
int T;
scanf("%d",&T);
int Case = 1;
while(T--)
{
int A, B, n;
scanf("%d %d %d",&A, &B, &n);
for(int i = 0; i < n; i++) scanf("%d",&f[i].a);
for(int i = 0; i < n; i++) scanf("%d",&f[i].b);
memset(dp, 0, sizeof(dp));
for(int i = 0; i < n; i++)
{
for(int j = A; j >= 0; j--)
{
if(j-f[i].a < 0) break;
dp[j] = max(dp[j], dp[j-f[i].a]+f[i].b);
}
}
printf("Case #%d: ",Case++);
if(dp[A] >= B) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
</span>
G:分大圓裡面,小圓裡面。分別推一下公式就可以了啊,小圓裡面我是先求交點又推的公式。
<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <time.h>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define read() freopen("data1.in", "r", stdin)
#define write() freopen("data1.out", "w", stdout);
#define mod 1000000007
const int maxn = 1000010;
char str[maxn];
using namespace std;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
LL n;
cin >>n;
if(!n)
{
cout<<2<<endl;
continue;
}
if(n == 1)
{
cout<<4<<endl;
continue;
}
LL y = 2*n;
y += (n+1)*n/2+1;
///if(n == 2) y++;
cout<<y<<endl;
}
}</span>
H:一個水題,結果錯了好多遍啊。讀題不認真啊。sad。。。
注意:i為偶數時i只能和i-1匹配。
<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <time.h>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define read() freopen("data1.in", "r", stdin)
#define write() freopen("data1.out", "w", stdout);
#define mod 1000000007
const int maxn = 1000010;
char str[maxn];
using namespace std;
bool judge(char x, char y)
{
if(x == 'a' && y == 'b' || x == 'b' && y == 'a') return 1;
if(x == 'c' && y == 'd' || x == 'd' && y == 'c') return 1;
if(x == 'e' && y == 'f' || x == 'f' && y == 'e') return 1;
if(x == 'g' && y == 'h' || x == 'h' && y == 'g') return 1;
if(x == 'i' && y == 'j' || x == 'j' && y == 'i') return 1;
if(x == 'k' && y == 'l' || x == 'l' && y == 'k') return 1;
if(x == 'm' && y == 'n' || x == 'n' && y == 'm') return 1;
if(x == 'o' && y == 'p' || x == 'p' && y == 'o') return 1;
if(x == 'q' && y == 'r' || x == 'r' && y == 'q') return 1;
if(x == 's' && y == 't' || x == 't' && y == 's') return 1;
if(x == 'u' && y == 'v' || x == 'v' && y == 'u') return 1;
if(x == 'w' && y == 'x' || x == 'x' && y == 'w') return 1;
if(x == 'y' && y == 'z' || x == 'z' && y == 'y') return 1;
return 0;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s",str);
int len = strlen(str);
char sx[maxn];
int top = 0;
for(int i = 0; i < len; i++)
{
if(!top)
{
sx[top++] = str[i];
continue;
}
if(abs(str[i]-sx[top-1]) == 1 && judge(str[i], sx[top-1])) top--;
else sx[top++] = str[i];
}
sx[top] = '\0';
if(!top)
{
puts("sad!");
continue;
}
puts(sx);
}
}</span>
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