結論 (自己未曾學到的知識補充)
以下幾乎全部抄的題解(方便自己復習)
因為這些我見都沒見過(_(:з」∠)_蒟蒻瑟瑟發抖
LuoGu P1887 乘積最大3
請你找出M個和為N的正整數,他們的乘積要盡可能的大。 輸出字典序最小的一種方案。
扔代碼
//#define fre yes #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; int n,m; template<typename T>inline void read(T&x) { x = 0;char c;int lenp = 1; do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c)); do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c)); x *= lenp; } int main () { read(n);read(m); int tot = n / m; for (int i=1;i<=m;i++) { if(i<m-(n-tot*m)+1) { printf("%d ",tot); } else printf("%d ",tot+1); } return 0; }
LuoGu P1075 質因數分解
一個數有且只能分解為一組質數的乘積
扔代碼
//#define fre yes #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; int k; template<typename T>inline void read(T&x) { x = 0;char c;int lenp = 1; do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c)); do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c)); x *= lenp; } int main() { read(k); for (int i=2;i<=k;i++) { if(k%i == 0) { printf("%d\n",k/i); return 0; } } return 0; }
LuoGu P3717 [AHOI2017初中組]cover
圓覆蓋問題..大概就是公式求圓覆蓋(兩點間的距離公式然後直到半徑好像也能求)
扔代碼
//#define fre yes #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 2005; int Map[maxn][maxn]; int n,m,r,ans; template<typename T>inline void read(T&x) { x = 0;char c;int lenp = 1; do { c = getchar();if(c == ‘-‘) lenp = - 1; } while(!isdigit(c)); do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c)); x *= lenp; } void kis(int x,int y,int z) { for (int i=1;i<=n;i++) { for (int j=1;j<=n;j++) { if(pow(i-x,2)+pow(j-y,2) <= pow(r,2)) { Map[i][j] = 1; } } } } int main () { read(n);read(m);read(r); for (int i=1;i<=m;i++) { int x,y; read(x);read(y); Map[x][y] = 1; kis(x,y,r); } for(int i=1;i<=n;i++) { for (int j=1;j<=n;j++) { if(Map[i][j] == 1) { ans++; } } } printf("%d\n",ans); return 0; }
LuoGu P2415 集合求和
給定一個集合s(集合元素數量<=30),求出此集合所有子集元素之和
樣例
輸入 2 3 輸出 10
解釋:
[] [2] [3] [2 3] 2+3+2+3=10
扔代碼
//#define fre yes
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 35;
int arr[maxn];
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main ()
{
int k = 1;
while(cin >> arr[k]) { k++; }
long long s = 0;
for (int i=1;i<=k;i++)
{
s += arr[i];
} s *= pow(2, k-2);
printf("%lld\n",s);
return 0;
}
LuoGu P1851 好朋友
相親數:兩個數中任意一個的所有因子(除本身外)之和等於另外一個數
扔代碼
//#define fre yes
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int pr[maxn];
int S;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
void cur()
{
// O(n^2)
// for(int i=1;i<=100000;i++)
// {
// for(int j=1;j<i;j++)
// {
// if(i%j==0)
// {
// pr[i]+=j;
// }
// }
// }
//O(n sqrt(n))
int j;
for (int i=1;i<=100000;i++)
{
for (j=1;j*j<=i;j++)
{
if(i%j==0)
{
pr[i] += j;
pr[i] += i/j;
}
} j--;
if(j*j == i) pr[i] -= j;
pr[i] -= i;
}
//O(n log(log(n)))
// for(int i=1;i<=20000;i++)
// for(int j=i;i*j<=20000;j++)
// pr[i*j]+=i;
}
int main()
{
read(S);
cur();
for (int i=S;;i++)
{
if(pr[pr[i]] == i&&pr[i]!=i)
{
printf("%d %d",i,pr[i]);
return 0;
}
} return 0;
}
LuoGu https://www.luogu.org/problemnew/show/P1548
有一個棋盤 給你n m 找有多少個正方形和長方形
扔代碼
//#define fre yes
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int ans1,ans2;
int n,m;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main ()
{
read(n);read(m);
ans2 = ((m+1)*(n+1)*m*n)/4;
for(int i=n,j=m;j>=1&&i>=1;j--,i--) {
ans1 += i*j;
} printf("%d %d\n",ans1,ans2-ans1);
return 0;
}
LuoGu P2181 對角線
凸多邊形找對角線交點個數
扔代碼
//#define fre yes
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
unsigned long long n;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main()
{
read(n);
n = (n*(n-1)/2*(n-2)/3*(n-3)/4);
printf("%lld\n",n);
return 0;
}
LuoGu P1469 找筷子
利用 a^a=0
//#define fre yes
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,ans;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1 ; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main()
{
read(n);
for (int i=1;i<=n;i++)
{
int x;
read(x);
ans = (ans ^ x);
} printf("%d\n",ans);
return 0;
}
LuoGu P1100 高低位交換
C++的32位無符號整形(話說有這個嗎 Orz)
左移16位,就是低位轉到高位;右移16位,就是高位轉到低位;兩者相加,就是新數
扔代碼
//#define fre yes
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main()
{
scanf("%u", &n);
printf("%u\n", (n >> 16) + (n << 16));
return 0;
}
LuoGu P1147 連續自然數和
設首項為L,末項為R,那麽sum(L,R)=(L+R)(R-L+1)/2=M
即(L+R)(R-L+1)=2M
可以把2M分解成兩個數之積,假設分成了兩個數K1,K2,且K1<K2時,
可以列一個二元一次方程組
R-L+1=K1
L+R=K2 解得L=(K2-K1+1)/2, R=(K1+K2-1)/2
當K1,K2一奇一偶時,L,R才有自然數解.
不過有一種特殊情況,就是L=R的情況,這種情況是不允許的
即(K2-K1+1)/2≠(K1+K2-1)/2,解得K1≠1
扔代碼
//#define fre yes
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main(){
read(n);
for(int i=sqrt(2*n);i>1;i--)
{
if(2*n%i==0 && (i+2*n/i)%2)
{
int j=2*n/i;
printf("%d %d\n",(j-i+1)/2,(i+j-1)/2);
}
} return 0;
}
LuoGu P1497 木牛流馬
讀了半天題都沒讀懂..但是題解寫的的確是結論
//#define fre yes
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,k,h;
long long ans = 1;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main()
{
read(n);read(k);read(h);
if(k > n)
{
puts("0");
return 0;
}
for (int i=1;i<=h;i++)
{
int x;
read(x);
for(int j=1;j<=x;j++)
ans *= j;
} long long ans_g = 1;
for (int i=n;i>=n-k+1;i--) ans_g *= i * i;
printf("%lld\n",ans_g/ans);
return 0;
}
LuoGu P2807 三角形計數
把大三角形的每條邊n等分,將對應的等分點連接起來
//#define fre yes
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
long long ans,n;
int k;
template<typename T>inline void read(T&x)
{
x = 0;char c;int lenp = 1;
do { c = getchar();if(c == ‘-‘) lenp = -1; } while(!isdigit(c));
do { x = x * 10 + c - ‘0‘;c = getchar(); } while(isdigit(c));
x *= lenp;
}
int main()
{
read(k);
while(k--)
{
ans = 0;
read(n);
for (long long i=0;n-i!=0;i++)
ans += (n-i) * (n-i+1);
for (long long i=0;n-i>=2;i+=2)
ans += (n-i) * (n-i-1);
ans /= 2;
printf("%lld\n",ans);
} return 0;
}
結論 (自己未曾學到的知識補充)