how to multiply in mips

/* Function: multiply a large integer * // * solution: read two large integers, put them in two strings respectively, and declare the third string to save the result. It involves character and number conversion techniques. For details, see the code. */# Include

Given a number 'n', find the smallest number 'P' such that if we multiply all digits of 'P', we get 'n '. the result 'P' shoshould have minimum two digits. Examples: Input: n = 36Output: p = 49 // Note that 4*9 = 36 and 49 is the smallest such numberInput: n = 100Output: p = 455// Note that 4*5*5 = 100 and 455 is the smallest such numberInput: n = 1Output:p = 11// Note that 1*1 = 1Input: n = 13Output: Not Possible For a given n, following are the tw

Problem Description: Quotient, cannot use multiplication, division, modulo operation.Algorithm idea: Can not use division, that can only use subtraction, but with subtraction, time-out. You can use the displacement operation, each divisor left shift, equivalent to twice times.1 Public classDividtwointegers {2 Public intDivideintDividend,intdivisor)3 {4 if(divisor = = 0)returnInteger.max_value;5 if(Divisor = =-1 dividend = = integer.min_value)returnInteger.max_value;6

contains the sequence F0, F1, ..., fn -1 ( C16>0≤ FI n) and the third-the sequence w0, w1, ..., wn -1 (0≤ wi ≤108). OutputPrint n lines, the pair of integers si, mi in each line. Examplesinput7 31 2 3 4 3 2 66 3 1 4 2 2 3Output10 18 17 110 28 27 19 3Input4 40 1 2 30 1 2 3Output0 04 18 212 3Input5 31 2 3) 4 04 1 2) 14 3Output7 117 119 221 38 1#include #include#include#include#include#defineFi first#defineSe Secondusing namespaceStd;typedefLong LongLl;typedef pairint>PII;Const intn=1e5+5;

selected effect; Used to determine the correspondence of curly braces; F2: Quickly navigate to the next error and warning, add Shift key, and navigate to the previous error; Ctrl+w the code by the syntax, the continuous press will have other effects, plus shift key, to produce a reverse selection effect Ctrl+alt+v suggest that the selection is a local variable Iv. deletion and insertion Ctrl+d: Copy the current line or copy the selection; Ctrl+y: Delete t

　　Add, subtract, multiply and divide functions for number type to solve the problem of the accuracy of the addition result of float

if(upp[i][u]!=1|| upp[i][v]!=-1)return 0; theU=par[i][u]; v=Par[i][v]; About } the } the if(upp[0][u]!=1|| upp[0][v]!=-1)return 0; the } + return 1; - } the Bayi intMain () { the intn,q,a,b; the while(~SCANF ("%d",N)) { -Ne=0; -memset (head,-1,sizeof(head)); the for(intI=1; ii) { thescanf"%d%d",a,b); theAddedge (a,b,-1); theAddedge (B,a,1); - } the theDfs1,1); the init (n);94 thescanf"%d",q); the while(q--)

#defineRJOP (n) for (int j=0;j#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong LongLL;//const LL mod=1e9+7;Const DoublePi=acos (-1.0);Const intinf=0x3f3f3f3f;Const intn=1e5+ -; LL X,m,k,c; LL MoD; ll Fastmod (ll X,ll y) {ll ans=1,Base=x; while(y) {if(y1) ans*=Base, ans%=MoD; Base*=Base; Base%=MoD; Y= (y>>1); } returnans;}intMain () {intT,cnt=1; scanf ("%d",t); while(t--) {printf ("Case #%d:\n", cnt++); scanf ("%i64d%i64d%i64d%i64d",x,m,k,c); MoD=9*K; LL FX=fastmod (Ten, M); LL ans= (

#include #includestring.h>#include#includeusing namespacestd;Const intmaxn= $;CharS1[MAXN],S2[MAXN];intA[MAXN],B[MAXN];int**ans=New int*[MAXN];intret[2* (maxn+1)];intMain () {intLen1,len2,i,j,k,k1,k2,num,sum,over; while(SCANF ("%s%s", s1,s2)! =EOF) {Len1=strlen (S1); Len2=strlen (S2); K1=0; for(i=len1-1; i>=0; i-=4) {sum=0; for(J=i; j>i-4j>=0; j--) {num=s1[j]-'0'; for(K=i-j; k>0; k--) Num*=Ten; Sum+=num; } a[k1++]=sum; } K2=0; for(i=len2-1; i>=0; i-=4) {sum=0; for(J=i; j>i-4j>=0; j--) {num

1#include 2#include 3#include 4#include 5 using namespacestd;6 intMain () {7 Chara[101],b[101];8 intans[ -],len_ans;9Cin>>a>>b;Ten One intLa=strlen (a); A intlb=strlen (b); - - int*na= (int*)malloc(la*sizeof(int)); the int*nb= (int*)malloc(lb*sizeof(int)); - - for(inti=la-1; i>=0; i--){ -na[i]=a[la-i-1]-'0'; + } - for(inti=lb-1; i>=0; i--){ +nb[i]=b[lb-i-1]-'0'; A } at -memset (ans,0,sizeof(ans)); - for(intI=0; i){ - for

Clc;clear All;close All; A = ' 962898798798798978987898 '; B = ' 9387978987987879899999999999999999999999999999999999999999999999999999999999999999 '; A = FLIPLR (a); b = FLIPLR (b); Result=zeros (1,length (a) +length (b)); For i = 1:length (b) multiflag =0;addflag =0;for II =1:length (a) Temp1 = s Tr2num (A (ii)) * Str2Num (B (i)) +multiflag;multiflag = Floor (TEMP1/10); temp1 = mod (temp1,10); Temp2 = result (i+ii-1) + Temp1 +addflag; Addflag = Floor (TEMP2/10); result (i+ii-1) = mod (temp2,10

Title Description: (link)Given numbers represented as strings, return multiplication of the numbers as a string.Note:the numbers can be arbitrarily large and is non-negative.Problem Solving Ideas:1 classSolution {2 Public:3 stringMultiplystringNUM1,stringnum2) {4 intLen1 =num1.size ();5 intLen2 =num2.size ();6 7 intLen = len1 +Len2;8 stringR (Len,'0');9 Ten for(inti =0; i i) { One intcarry =0; A intA = Num1[len

#include 　　[String] Multiply trees

Because of the reasons for the need to do so, no nonsense, directly on the code, I use the GridView binding data, table is the same, because the GridView code compiled by the browser is the table. The following is the HTML code for the ASPX page:Width= "100%" allowpaging= "True" pagesize= "emptydatatext=" did not find the relevant data! "Onpageindexchanging=" > "gv_new_pageindexchanging"The next step is the jquery code:Don't forget to quote jquery Kochhar, I didn't write it out here.Summary: I w

Counting order ExpectationsRujia Teacher's White book on the example ... See Petition1 //UVA 110212#include 3#include 4 #defineRep (i,n) for (int i=0;i5 #defineF (i,j,n) for (int i=j;i6 Const intn=1010;7 intn,m,k;8 DoubleP[n],f[n];9 intMain () {Ten intT; Onescanf"%d",t); AF (CS,1, T) { -scanf"%d%d%d",n,k,m); -Rep (i,n) scanf ("%LF",p[i]); thef[0]=0; f[1]=p[0]; -F (I,2, M) { -f[i]=0; -Rep (j,n) F[i]+=p[j]*pow (f[i-1],j); + } -printf"Case #%d:%.7lf\n", Cs,pow (f[m],k)); + } A r

Advertising:#include int main(){ puts("转载请注明出处[vmurder]谢谢"); puts("网址：blog.csdn.net/vmurder/article/details/44015679");}ExercisesIn fact, no matter, just put a template + understanding commentsCode:#include #include #include #include #include #include using namespace STD;#define N 131075intN,c[n];ComplexDouble> A[n],b[n],p[n];Const DoublePi=ACOs(-1);voidFFT (ComplexDouble> x[],intNintType) {intI,j,k,t;/* Press bits to flip: for example 01234567 04261537 000 000 001 100 010 010 011 110 100

modulus, division to seek the inverse of the yuan, where the expansion of Euclid easy to find.1#include 2#include 3#include 4#include 5#include 6 #defineLL Long Long7 #defineMoD 10000000078 #defineIn Freopen ("In.txt", "R", stdin);9 using namespacestd;Ten One LL x,y,gcd; A voidEX_GCD (LL a,ll b) - { - if(!B) {x=1; y=0; gcd=A;} the Else{EX_GCD (b,a%b); LL temp=x;x=y;y=temp-a/b*y;} - } - - intMainintargcChar Const*argv[]) + { - //in ; + A LL two,six; atEX_GCD (2, MoD); - while

(b[foot2[x]]-b[x)); x:=foot2[x];End;End;beginREADLN (n); fori:=1 toN Do beginRead (A[i]); G[i]:=i;End; b:=A; Qsort (1, N); Done Make READLN (t); b[0]:=-maxlongint; ans:=0; ans1:=1; ans2:=0; fori:=1 toN Do beginS1:=0; s2:=0; Find (I,t); ifS1*ans2 Then beginAns:=i; ANS1:=S1; ANS2:=S2;End Else if(S1*ANS2=S2*ANS1) and( not((s1=0) and(s2=0))) Then ifB[i]>b[ans] Then beginAns:=i; ANS1:=S1; ANS2:=S2;End End; Writeln (ANS); READLN (m); fori:=1 toM Do beginreadln (x,t); S1:=0

#Lang Racket (define (fast-multiplication a b N); A*n (Cond (= n0) b); n==0( even N) (Fast-multiplication (Doublea) b (halve n)); even? N (Else(Fast-multiplication A (+ a B) (-N1)) );Else); cond); Fast-multiplication it's Iteration(define (fast-multiplication-Newa N) (Cond (= n0)0) ((= N1) a) ((even n) (Double(fast-multiplication-NewA (halve N)))) (Else(A + (fast-multiplication-NewA (-N1))) );Else); cond); fast-multiplication-NewIt's Recursive(Define (Doublex) (+ x x));Double(define (Hal

Given numbers represented as strings, return multiplication of the numbers as a string.Note:the numbers can be arbitrarily large and is non-negative.Subscribe to see which companies asked this questionA large number multiplication algorithm used in ACM contests, which uses strings to represent large numbers for calculation.stringMultiplystringNUM1,stringnum2) { stringNum (num1.size () + num2.size (),'0'); for(inti = num1.size ()-1; I >=0; --i) {intcarry =0; for(intj = num2.size ()-1; J >=0;