The number of n in a set of numbers, so that the sum of n equals a given value M

"topic analysis" in a set of numbers whether there is n number, so that the sum of n is equal to a given value m.

"Thinking Analysis": Analysis of data in the case:

1): 0 < n <= 100000

2): 0 < sequence length <= 100000

3): 0 <= m <=100000

Data is small: with a similar idea of DP,

Code:

#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream>using namespace STD; #define REP (i,j,k) for (int i= (int) j;i< (int.) k;++i) #define PER (i,j,k) for (int i= (int) j;i> (int) k;--i) #define Lowbit (a) A&-a#define Max (A, B) a>b?a:b#define Min (A, B) a>b?b:a#define mem (A, B) memset (A,b,sizeof (a)) typedef Long long ll;typedef unsigned long long llu;typedef double db;const int n=1e5+10;const int inf=0x3f3f3f3f;int t,n,m;int PR    E[n],last[n];int Num[n];int Main () {int x=0;    Char ch;    int sum=1;    MEM (last,0);        while (~ (Ch=getchar ())) {if (IsDigit (CH)) x=x*10+ch-' 0 ';            else if (ch!= ' \ n ') {pre[sum++]=x;        x=0;            } else{pre[sum++]=x;            MEM (last,0);            Last[0]=1;            M=read ();                    for (int i=1, i<sum; ++i) {for (int j=m; j>=0;--j) {int k=pre[i]+j;                if (last[j]&&k<=m) last[k]=1;    cout<< "k=" <<k<< "<<" Last[k "<<last[k]<<endl;            }} if (Last[m]) puts ("Yes");            Else puts ("No");            x=0;            Sum=1;        MEM (last,0); }} return 0;}

The number of n in a set of numbers, so that the sum of n equals a given value M