Problem A King Moves
题目大意
有一个国际象棋的棋盘，给定一个国王的位置，求其移动一步可到达的合法位置数量。
解题分析
国王的位置可以分为3类，每类的答案为8、5、3。
参考程序
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <string> 8 #include <vector> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <cassert> 13 #include <iostream> 14 #include <algorithm> 15 #pragma comment(linker,"/STACK:102400000,102400000") 16 using namespace std; 17 18 #define N 508 19 #define M 50008 20 #define LL long long 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 #define clr(x,v) memset(x,v,sizeof(x)); 24 #define bitcnt(x) __builtin_popcount(x) 25 #define rep(x,y,z) for (int x=y;x<=z;x++) 26 #define repd(x,y,z) for (int x=y;x>=z;x--) 27 const int mo = 1000000007; 28 const int inf = 0x3f3f3f3f; 29 const int INF = 2000000000; 30 /**************************************************************************/ 31 32 int main(){ 33 char c,n; 34 scanf("%c%c",&c,&n); 35 int num=0; 36 if (c=='a' || c=='h') num++; 37 if (n=='1' || n=='8') num++; 38 if (num==0) puts("8"); 39 if (num==1) puts("5"); 40 if (num==2) puts("3"); 41 }
Problem B Optimal Point on a Line
题目大意
在一条直线上有n个点，求某个点使得其到其他点的距离之和最小。
解题分析
排序之后输出正中间的点即可。
参考程序
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <string> 8 #include <vector> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <cassert> 13 #include <iostream> 14 #include <algorithm> 15 #pragma comment(linker,"/STACK:102400000,102400000") 16 using namespace std; 17 18 #define N 508 19 #define M 50008 20 #define LL long long 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 #define clr(x,v) memset(x,v,sizeof(x)); 24 #define bitcnt(x) __builtin_popcount(x) 25 #define rep(x,y,z) for (int x=y;x<=z;x++) 26 #define repd(x,y,z) for (int x=y;x>=z;x--) 27 const int mo = 1000000007; 28 const int inf = 0x3f3f3f3f; 29 const int INF = 2000000000; 30 /**************************************************************************/ 31 int n; 32 int a[300008]; 33 int main(){ 34 scanf("%d",&n); 35 rep(i,1,n) scanf("%d",&a[i]); 36 sort(a+1,a+n+1); 37 printf("%d ",a[(n+1)/2]); 38 }
Problem C Magic Odd Square
题目大意
将1~n^2的数字填入n*n的矩形，使得每行每列每条正对角线的数字之和均为奇数。
n <= 49 , 且n为奇数。
解题分析
通过观察样例和探索后可以发现一种合法的构造方案。
在n*n的矩形正中间放置一个完全由奇数组成的旋转45度的（n-1）*（n-1）的矩形。
参考程序
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <string> 8 #include <vector> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <cassert> 13 #include <iostream> 14 #include <algorithm> 15 #pragma comment(linker,"/STACK:102400000,102400000") 16 using namespace std; 17 18 #define N 508 19 #define M 50008 20 #define LL long long 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 #define clr(x,v) memset(x,v,sizeof(x)); 24 #define bitcnt(x) __builtin_popcount(x) 25 #define rep(x,y,z) for (int x=y;x<=z;x++) 26 #define repd(x,y,z) for (int x=y;x>=z;x--) 27 const int mo = 1000000007; 28 const int inf = 0x3f3f3f3f; 29 const int INF = 2000000000; 30 /**************************************************************************/ 31 int n,x,y; 32 int a[108][108]; 33 int main(){ 34 scanf("%d",&n); 35 int mid=(n+1)/2; 36 rep(i,1,mid){ 37 rep(j,mid-i+1,mid+i-1) 38 a[i][j]=1; 39 } 40 rep(i,1,mid-1){ 41 rep(j,mid-i+1,mid+i-1) 42 a[n-i+1][j]=1; 43 } 44 x=1; y=2; 45 rep(i,1,n) 46 { 47 rep(j,1,n) 48 if (a[i][j]==1) 49 { 50 printf("%d ",x); 51 x+=2; 52 } 53 else 54 { 55 printf("%d ",y); 56 y+=2; 57 } 58 printf(" "); 59 } 60 61 }
Problem D Two Arithmetic Progressions
题目大意
给定a1, b1, a2, b2, L, R (0 < a1, a2 ≤ 2·109, - 2·109 ≤ b1, b2, L, R ≤ 2·109, L ≤ R)。
求所有满足条件的x的数量，x=a1*k+b1=a2*l+b2,且 L≤x≤R。
解题分析
x=a1*k'+b1=a2*l'+b2
移项后可得 a1*k-a2*l=b2-b1。
用exgcd解出最小的k',l'。
则k=k'+(a2/gcd(a1,a2))*m。
由 k>=0 ，l>=0 和 L<=a1*k+b1<=R 解出m的范围。
参考程序
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <string> 8 #include <vector> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <cassert> 13 #include <iostream> 14 #include <algorithm> 15 #pragma comment(linker,"/STACK:102400000,102400000") 16 using namespace std; 17 18 #define N 508 19 #define M 50008 20 #define LL long long 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 #define clr(x,v) memset(x,v,sizeof(x)); 24 #define bitcnt(x) __builtin_popcount(x) 25 #define rep(x,y,z) for (int x=y;x<=z;x++) 26 #define repd(x,y,z) for (int x=y;x>=z;x--) 27 const int mo = 1000000007; 28 const int inf = 0x3f3f3f3f; 29 const LL INF = 20000000000000ll; 30 /**************************************************************************/ 31 32 LL a1,b1,a2,b2,L,R,x,y,d; 33 34 LL gcd(LL x,LL y){ 35 return y == 0 ? x : gcd(y,x % y); 36 } 37 38 LL exgcd(LL a,LL b,LL &x,LL &y){ 39 if (b==0){ 40 x=1; 41 y=0; 42 return a; 43 } 44 LL d=exgcd(b,a%b,y,x); 45 y-=x*(a/b); 46 return d; 47 } 48 LL up(LL x,LL y){ 49 if (x>0) return x % y ? x / y + 1 : x / y; 50 return x / y; 51 } 52 LL down(LL x,LL y){ 53 if (x>0) return x / y; 54 return x % y ? x / y - 1 : x / y; 55 } 56 int main(){ 57 scanf("%lld%lld%lld%lld%lld%lld",&a1,&b1,&a2,&b2,&L,&R); 58 d=exgcd(a1,-a2,x,y); 59 if ((b2-b1) % d!=0){ 60 printf("%d ",0); 61 return 0; 62 } 63 LL k0 = x * (b2-b1) / d; 64 k0 =(k0 % (a2/d) + (a2/d)) % (a2/d); 65 LL l0 = (a1*k0+b1-b2) /a2; 66 67 // cout << k0 << " " << l0 << endl; 68 LL l,r,ll,lll,rr,rrr; 69 if (d>0) 70 { 71 r = down((R-a1*k0-b1)*d,a1*a2); 72 l = up((L-a1*k0-b1)*d,a1*a2); 73 ll = up(-k0*d,a2); 74 lll = up(-l0*d,a1); 75 l=max(max(ll,lll),l); 76 } 77 else 78 { 79 l = up((R-a1*k0-b1)*d,a1*a2); 80 r = down((L-a1*k0-b1)*d,a1*a2); 81 rr = down(-k0*d,a2); 82 rrr = down(-l0*d,a1); 83 r=min(min(rr,rrr),r); 84 85 LL x1 = (R-a1*k0-b1)*d ; 86 LL x2 = (L-a1*k0-b1)*d ; 87 LL y1 = a1 * a2; 88 89 } 90 printf("%lld ",r-l+1>0 ? r-l+1 : 0 ); 91 }
Problem E Generate a String
题目大意
每次操作可以花费x的代价将数字加1或减1，或者花费y的代价将数字乘2。
求将数字从0变到数字n的最小代价。
n<=10^7。
解题分析
可以发现一个显然的dp方程
dp[i]=min{dp[i-1]+x,dp[i+1]+x,dp[i/2]+y(2|i)}
由于n很大，不能高斯消元。
经过思考后可以发现-1的操作不会连续执行两次（必然可以通过+1和*2来代替）。
所以每次dp[i]时，顺便求出dp[i+1],再由dp[i+1]来更新dp[i]。
参考程序
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <string> 8 #include <vector> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <cassert> 13 #include <iostream> 14 #include <algorithm> 15 #pragma comment(linker,"/STACK:102400000,102400000") 16 using namespace std; 17 18 #define N 508 19 #define M 50008 20 #define LL long long 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 #define clr(x,v) memset(x,v,sizeof(x)); 24 #define bitcnt(x) __builtin_popcount(x) 25 #define rep(x,y,z) for (int x=y;x<=z;x++) 26 #define repd(x,y,z) for (int x=y;x>=z;x--) 27 const int mo = 1000000007; 28 const int inf = 0x3f3f3f3f; 29 const int INF = 2000000000; 30 /**************************************************************************/ 31 32 LL f[10000008]; 33 34 int n,x,y; 35 36 int main(){ 37 scanf("%d%d%d",&n,&x,&y); 38 f[0]=0; 39 rep(i,1,n){ 40 f[i]=f[i-1]+x; 41 if (i % 2 == 0) f[i]=min(f[i],f[i/2]+y); 42 43 f[i+1]=f[i]+x; 44 if ((i+1) % 2==0) f[i+1]=min(f[i+1],f[(i+1)/2]+y); 45 46 f[i]=min(f[i],f[i+1]+x); 47 48 } 49 printf("%I64d ",f[n]); 50 }
Problem F String Set Queries