ACM integer Division (iv)

Integer Division (iv) time limit:MS | Memory limit:65535 KB Difficulty:3

Describe

Summer vacation came, HRDV again to stay in the school to participate in ACM training, Training Life very happy (PS: you know), but he recently encountered a problem, let him baffled his solution, he was very depressed. Honey, can you help him?

The problem is that we often see the integer division, given two integers n, m, required to add m-1 multiplication sign in n, divide n into m segments, and find the maximum product of this m segment

Input

The first line is an integer t, which indicates that there is a T group of test data
Next T line, each line has two positive integers n,m (1<= n < 10^19, 0 < m <= n digits);

Output

output Each set of test sample results as an integer in one row

Sample input

2111 21111 2

Sample output

11121


Using the idea of dynamic planning, thought for a whole afternoon, did not expect to finally make a 5-layer cycle of dynamic planning. Fortunately, the topic of AC, and then see the answer on the Internet, using the DP interval dynamic programming algorithm, today do not study, some tired. Here's my algorithm.

#include <string>#include<sstream>#include<iostream>#include<string.h>#include<stdio.h>using namespacestd;#defineNUM 20Long LongAtoll (Const Char*p)    {StringStream strvalue; strvalue<<p; Long Longvalue; strvalue>>value; returnvalue;}intMain () {intcount; scanf ("%d", &count);  while(count--)    {        intI,j,k,l,m,n; Long LongMm[num][num][num]; CharNum[num]; CharTemp[num]; scanf ("%s", num); N=strlen (num); scanf ("%d", &m);  for(k =0; K < M; k++)        {             for(i =0; I < n; i++)            {                 for(j = i + K; j < N; j + +)                {                    if(k = =0) {strncpy (temp, num+i, j-i+1); Temp[j-i+1] =' /'; MM[I][J][K]=Atoll (temp); }                    Else                    {                        Long Longres =-1;  for(L =0; L < J-i; l++)                        {                            intO, k1=0, k2=0;  for(o =0; o < L +1&& O < K; o++) {K1=o; K2= K-1-K1; Long LongTRes = mm[i][i+l][k1] * mm[i+l+1][J][K2]; Res= res > TRes?Res:tres; }} Mm[i][j][k]=Res; } }}} cout<< mm[0][n-1][m-1] <<Endl; }    return 0;}

DP interval algorithm

#include <iostream>#include<cstdio>#include<string>#include<cmath>#include<algorithm>typedefLong LongLl;typedefLong Longarray[ A]; Array ob;ll n,m,res,reco,rec,cur,maxs,temp;using namespaceStd;ll Pow_dfs (ll i) {ll ress=1;  for(LL j=1; j<=i;j++) {ress*=Ten; }    returnress;} ll Maxmin (ll A,ll b) {returnA>b?a:b;}voidDFS (ll pos,ll rec,ll now) {if(rec==m-1) {Ob[rec]=Now ; Cur=1;  for(intI=0; i<m;i++) {cur*=Ob[i]; } maxs=maxmin (CUR,MAXS); return ; }    if(pos>=reco)return ; DFS (POS+1, Rec,now); OB[REC]=now/pow_dfs (reco-POS); now=now%pow_dfs (reco-POS); DFS (POS+1, rec+1, now);}intMain () {ll T; //freopen ("D://imput.txt "," R ", stdin);scanf"%lld",&T);  while(t--) {Reco=1; maxs=0; scanf ("%lld%lld",&n,&m); Temp=N;  while(temp/Ten) {Temp=temp/Ten; Reco++; } DFS (1,0, N); printf ("%lld", MAXS); if(t!=0) printf ("\ n"); }    return 0;}

ACM integer Division (iv)