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1. pseudo_code: find_median(m1, n1, m2, n2) { if(n1-m1<=2 or n2-m2<=2) { find and print median from “query(m1, D1), … ,querr(n1,D1), query(m2, D2), … ,querr(n2, D2),”; return; } if(quer

y((n1+m1)/2, D1) > query((n2+m2)/2, D2)) find_median(m1,n1+m1)/2,(n2+m2)/2, n2); else find_median((n1+m1)/2,n1, m2, (n2+m2)/2); } explanation:m1, n1 is range of medain in D1, if we find median_one of D1 and median_two of D2, we can delete the part above a larger number of both medians and the part under a smaller number of both medians. subproblem reduction graph:

Analyse the complexity of the algorithm: O(log n *2 +2) = O(log n) Prove the correctness of the answer: first, every substep is right. because the median of 2n number can’t exist in either the part above a larger number of the medians or the part under a smaller number of the medians. So deleting them is right. And then we prove that the median of 2n equal to the median of subproblem. two part of deleting is equivalent and if we delete the same numbers on either side of the median, the median is not changed. above of all, my algorithm is right. by the way, deleting should be same numbers in algorithm implementation.

4. pseudo_code: find_local_min(root) { if root have no son return root; if probe(root)>probe(left) find_local_min(left); else if probe(root)>probe(right) find_local_min(right); else return root; } subproblem reduction graph:

Analyse the complexity of the algorithm: if c= 1 / and 4 ≥ ci≥ 1,c 2 ≤42 . sothe complexity of the algorithm is O(42 )= O(2 ). Prove the correctness of the answer: if there is a local_min node in the path from root to leaf.it will be found out by the algorithm. but, if not, the leaf is smaller than its father, so the leaf must be local_min node. so whatever, a local_min node can be found out.

5.algorithm: first, probe all node in six line including horizontal line 1, horizontal line n, horizontal line (n+1)/2, vertical line 1. vertical line n. vertical line (n+1)/2. Six line divide into four part. Then find the min node on the six line, and judge whether the minimum node is local_min node. if is, return; if not, node there is smaller node n around the minimum, so keep finding by using recursive function.from the part including the noden .

subproblem reduction graph:

Analyse the complexity of the algorithm: according to master theorem, if T n = T
n 4

+ , T(n)=O(n)

Prove the correctness of the answer: assumpt my answer is wrong, so there is no local_min node in sub part. n in the part is not local_min node, so there is a node n <n , obviously n is smaller than four line all around the part. every node beside the part is smaller than one node labled. so there is no minimum node in the part. however, Every node is distinct, there must be minimum node, so the assumption is wrong. my answer is right.

7. i)Code: #include <stdio.h> #include<stdlib.h> int N=100000; int data[100000]; long long merge_and_count(int l,int m,int r) { int count = 0; int* L = (int*)malloc((m-l+1)*sizeof(int)); int* R = (int*)malloc((r-m)*sizeof(int)); for(int i=l; i<=m; i++) L[i-l] = data[i]; for(int i=m+1; i<=r; i++) R[i-m-1] = data[i]; int i = 0,j = 0,k = l; for(i=0; i<m-l+1 && j<r-m;) { if(L[i] > R[j]) { data[k++]=R[j++]; count+=m-l-i+1; } else data[k++]=R[i++]; } for(;i<m-l+1;) data[k++]=L[i++]; for(;j<r-m;) data[k++]=R[j++]; return count; } long long inversion_Count(int l,int r) { FILE* fp = fopen("res.txt","a"); if(l < r) { int m=(l+r)>>1; long long L=inversion_Count(l,m); long long R=inversion_Count(m+1,r); long long LR=merge_and_count(l,m,r); long long tmp = L+R+LR; fwrite(&tmp, sizeof(long long), 1, fp); fclose(fp);

return tmp; } else { fclose(fp); return 0; } } int main() { FILE* fp = fopen("Q5.txt","r"); for(int i=0;i<N;i++) fread(&data[i], sizeof(int), 1, fp); printf("%lld",inversion_Count(0,N-1)); fclose(fp); return 0; } The result is 2951724343. ii) No. Because there is not the sorted half array. in fact, we get two sorted half array and can compute one number in left array and another number in right array by uing Merge_Sort. But Quick_Sort function is the same with visiting tree in preorder. we can’t insert same steps into the later.

10. karatsuba algorithm code: long long karatsuba(long long x, long long y, long long dnum) { if (x<10 || y<10) return x*y; long long xh = x, xl = 0, xt = 0; long long yh = y, yl = 0, yt = 0; for (int i=0;i<dnum/2;i++) { xt = xh % 10; xl = xl * 10 + xt; xh = xh / 10; } for (int i=0;i<dnum/2;i++) { yt = yh % 10; yl = yl * 10 + yt;

yh = yh / 10; } long long k0 = karatsuba(xl,yl,dnum/2); long long k1 = karatsuba((xl+xh), (yl+yh),dnum/2); long long k2 = karatsuba(xh,yh,dnum/2); long long tmp = k1-k2-k0; for (int i=0; i<dnum/2;i++) { tmp=tmp*10; } for (int i=0; i<dnum;i++) { k2=k2*10; } return k2+tmp+k0; }

quadratic grade-school method code. long long Multiplication(long long x, long long y, long long dnum) { long long xp[MAXDNUM]; long long yp[MAXDNUM]; long long xt = 0, yt = 0; for (int i=0;i<dnum;i++) { xt = x % 10; xp[i]=xt; x = x / 10; yt = y % 10; yp[i] = yt; y = y / 10; } long long s = 0; long long s1=0; for (int i=0;i<dnum;i++) { for (int j=0;j<dnum;j++) { long long tmp = 1;
for (int k = 0;k<j;k++) { tmp = tmp*10; }

s1= s1 + xp[i] * yp[j] * tmp; } long long tmp = 1; for (int k = 0;k<i;k++) { tmp = tmp*10; } s =s+s1*tmp; s1 = 0; } return s; }

compare the performance(/um) digit number quadratic grade-school method 2 ~ 4 1 8 3 16 17 (intel i5 2.5GHz, 4G memory, win7)

karatsuba algorithm ~ 1 4 20


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