mips multiply

is to divide an n-side shape into n-2 triangles and add them.And then ask for the area of each triangleSet a triangle of three points x1,y1,x2,y2,x3. Press counterclockwise.Then the area of the triangle can be expressed as s= ((x2-x1) * (y3-y1)-(x3-x1) * (Y2-Y1))/2;Put code on it.#include HDU 2036 | | nyoj1011 Fork Multiply Polygon area (both concave and convex edges can be)

Code:1#include 2#include string>3#include 4 5 using namespacestd;6 7 voidSwapChara,Charb)8 {9 Charc =A;TenA =b; Oneb =C; A } - - voidReversestrings) the { - intL =s.size (); - inti =0; - while(I 1) /2) + { -Swap (S[i], s[l-i-1]); +i++; A } at } - - stringMultiplystringNUM1,stringnum2) - { - if(NUM1 = ="0"|| Num2 = ="0") - return "0"; in intL1 =num1.length (); - intL2 =num2.length (); to reverse (NUM1); + reverse (num2); - int*result =New int[l1+

n1[s1.size ()] (this is, of course, a dynamic array when the code is actually written, here for convenience). The array that is converted to two arrays is multiplied by a phase. That is, the array N1 and the N2 are multiplied, saved in the arrays N3, it is important to note that the results of the multiplication of n1[i] and N2[j] are saved in n3[i+j]. The numbers in the array N3 are processed to carry the numbers that require rounding. For example, the number in N3[i] retains the value

Large number structure, tag can be put (sign)Wei can be put in lengthNum puts one digit in the valueNext put down one (higher one)public class bignum{Boolean tag = true;int num = 0;int wei= 1;Bignum Next;}public class Shuju {public static void Main (string[] args) {TODO auto-generated Method StubBignum a = new Bignum ();Getdemo (a);System.out.println ("---------a------------------");Out (a);Bignum B = new Bignum ();Getdemo (b);System.out.println ("-----------b---------------");Out (b);Bignum Big

Tag:uva348 optimal matrix link classic interval dp uva348 optimal matrix chain multiplication//typical interval dp//dp[i][j] represents the minimum cost of the matrix I to J chain multiplication//dp[i][j] = min (dp[i][k]+dp[k+1][j]+a[i].pl*a[k].pr*a[j].pr) ;//On the interval I to J find a K make dp[i][k]+dp[k+1][j] these two parts and in addition to the last//A[I].PL*A[K].PR*P[I].PR minimum value;//can have such a state key is; P =a[1] * a[2] * .... * A[k]//And q= a[k+1] * a[k+2] * .... * A[n]

#include #include#include#include#defineLL Long Long#defineMOD 23333333333using namespacestd; LL n,m; ll Qmul (ll A, ll B, ll c) {ll ans=0; A= a%C; while(b) {if(b1) ans = (ans+a)%C; b>>=1; A= (a+a)%C; } returnans;}intMain () { while(~SCANF ("%lld%lld",n,m)) {LL x= m/3, y = m%3, ans =0; if(%2) ans = qmul (x*3,(2*n+x+1)/2, MOD); ElseAns = Qmul (x*3/2,2*n+x+1, MOD); for(inti =0; i MOD; printf ("%lld\n", ans); } return 0;}//XDOJ1182Fast Multiply m

http://acm.xidian.edu.cn/problem.php?id=1182Similar to the fast power, the multiplication of the modulus of super long Long is handled.#include #include#include#include#defineLL Long Long#defineMOD 23333333333using namespacestd; LL n,m; ll Qmul (ll A, ll B, ll c) {ll ans=0; A= a%C; while(b) {if(b1) ans = (ans+a)%C; b>>=1; A= (a+a)%C; } returnans;}intMain () { while(~SCANF ("%lld%lld",n,m)) {LL x= m/3, y = m%3, ans =0; if(%2) ans = qmul (x*3,(2*n+x+1)/2, MOD); ElseAns = Qmul (x*3/2,2*n+x+1, M

Multiply game Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 1450 accepted submission (s): 508 Problem descriptiontired of playing computer games, alpc23 is planning to play a game on numbers. because plus and Subtraction is too easy for this gay, he wants to do some multiplication in a number sequence. after playing it a few times, he has found it is also too boring. so he plan to do a more chal

HDU-3074-Multiply game Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 3074 Evaluate the product of the range element. You can update the element and the line segment tree. # Include

The main idea is to ask for the shortest path from S to T of K-steps.Idea: Set F[P][I][J] for the shortest path from I to J just walk 2^p step, the DP equation is very simple:f[p][i][j] = min (F[p][i][j],f[p-1][i][k] + f[p-1][k][j]);The total number of steps T binary split, in T has 1 position, if this position is P, then use f[p][][] to update the answer g[][], and finally get the g[][] is the answer matrix.Attention to Discretization:CODE:#include Bzoj 1706 Usaco Relays Cow relay run/POJ 3613

For a given forest, each vertex has the right value and provides two operations: 1. query the K-Weight Value in the path between two points 2. Connect an edge between the two points to ensure the connection is still a forest The persistent part of the line segment tree is the same as the count on a tree, but this question adds a join operation. For connection operations, we need to use heuristic merge to reconstruct the small tree. This is a simple operation, but it can be proved that it is a di

/** * Seek the powers of a number * x^y,y is a positive integer.The calculator can only calculate two number multiplication, cannot calculate n number multiplication at one time.* Know: 2^5= (2^2) ^2*2; 2^6= (2^2) ^3= ((4) ^2); 2^8= (2^2) ^4= (4^2) ^2= 16^2 * Get the rule: x^y= (x^2) ^ (Y/2), define A=X^2,B=Y/2, then get shape such as: x^y= a^b; * y assumes an odd number, then the decomposition of the last multiplied by a (as above 2^6 decomposition into 4^3): x^y=a^b*x. * * Recursiv

Package Com.learn.algorithm.ploy;import Java.util.linkedlist;import Java.util.list;import java.util.Scanner;/** * Polynomial-related Operations */public class Ploy {public static void main (string[] args) {listIt can also be implemented in maps, and it's easierUsing the Java List to implement polynomial addition, multiply

Usually our website will load some JS code, statistics ah, Google ads Ah, Baidu League ah, Ali mom Advertising code Ah,A bunch, and finally make the page load slowly, very slow.Workaround: Change the way JS is included to allow JavaScript to load exponentially.Put the usual load mode:Change into:We can put it in this way to load (still on the bottom of the page):If this is "bad mood", it will naturally not load the JS file you do not want.How to multiply