POJ Dollar Dayz USD Holiday (full backpack, regular + large number)

Test instructions: Given the amount of money owned by the integers n and k,n, K represents a K tool, the price of which is 1~k. Find out how many ways to buy the N-ary.

Idea: K maximum 100, the number is too large, to use a large number. Other basic and complete backpacks are the same.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5 using namespacestd;6 Const intb=1005, n= Max;7 intcoin[ -], N, K;8 9 structbign {//please set the constant n as the length of the buffer and the maximum integer first. Ten     intlen, sex; One     intS[n]; A  -Bign () {//Create a new large number -          ThisLen =1; the          ThisSex =0; -memset (s),0,sizeof(s)); -     } -Bign (intNumber) {* This= number;}//creates a large number, initialized with an integral type +Bign (Const Char* Number) {* This= number;}//creates a large number, initialized with a string -  +     void operator= (Const Char*number) {//overload equals equal to a string A         intBegin =0; atLen =0; -Sex =1; -         if(Number[begin] = ='-') { -Sex =-1; -begin++; -         } in         Else if(Number[begin] = ='+') -begin++; to  +          for(intj = begin; NUMBER[J]; J + +) -s[len++] = number[j]-'0'; the     } *  $Bignoperator= (intNumber) {//overload equals equal to an integral typePanax Notoginseng         CharStr[n];//notice here that n, to set a buffer to be used outside.  -sprintf (str,"%d", number); the* This=str; +         return* This; A     } the  + bign Change (bign cur) { - bign now; $now =cur; $          for(inti =0; i < Cur.len; i++) -Now.s[i] = cur.s[cur.len-i-1]; -         returnNow ; the     } - Wuyi     voidDelzore () {//remove the leading 0. theBign now = Change (* This); -          while(Now.s[now.len-1] ==0&& Now.len >1) { Wunow.len--; -         } About* This=Change (now); $     } -  -     voidPut () {//The output value.  - Delzore (); A         if(Sex <0&& (len! =1|| s[0] !=0)) +cout <<"-"; the          for(inti =0; i < Len; i++) -cout <<S[i]; $     } the  theBignoperator+ (ConstBign &cur) {//can only be large number operator large number the bign sum, a, B; theSum.len =0; -A = A.change (* This); inb =b.change (cur); the  the          for(inti =0, G =0; G | | I < A.len | | i < B.len; i++){ About             intx =G; the             if(I < A.len) x + =A.s[i]; the             if(I < B.len) x + =B.s[i]; thesum.s[sum.len++] = x%Ten; +g = x/Ten; -         } the         returnsum.change (sum);Bayi     } the  theBignoperator- (ConstBign &cur) {//can only be large number operator large number - bign sum, a, B; -Sum.len =Len; theA = A.change (* This); theb =b.change (cur); the  the          for(inti =0; i < B.len; i++) { -Sum.s[i] = a.s[i]-b.s[i] +Sum.s[i]; the             if(Sum.s[i] <0) { theSum.s[i] + =Ten; theSum.s[i +1]--;94             } the         } the          for(inti = B.len; i < A.len; i++) { theSum.s[i] + =A.s[i];98             if(Sum.s[i] <0) { AboutSum.s[i] + =Ten; -Sum.s[i +1]--;101             }102         }103         returnsum.change (sum);104     } the };106 107 voidcal ()108 {109dp[0]=1; the      for(intI=1; i<=n; i++) dp[i]=0;//Initialize111     intTmp=min (K,n);//K>n parts can't afford to buy the      for(intI=0; i<tmp; i++)113          for(intj=0; j+coin[i]<=n; J + +) theDp[j+coin[i]] = Dp[j+coin[i]] +Dp[j]; the  the dp[n].put ();117 }118 119 intMain () { -     //freopen ("Input.txt", "R", stdin);121      for(intI=0; i< -; i++) coin[i]=i+1;//Initialize122      while(~SCANF ("%d%d",&n,&k))123 cal ();124     return 0; the}

AC Code

POJ Dollar Dayz USD Holiday (full backpack, regular + large number)