退背包（好像是这么叫的）

\(f[j]\)是正常的01背包统计方案数，\(g[i][j]\)是删去第i个物品后的01背包统计方案数

#include
    
     #include
     
      #include
      
       using namespace std; inline int rd(){    int ret=0,f=1;char c;    while(c=getchar(),!isdigit(c))f=c=='-'?-1:1;    while(isdigit(c))ret=ret*10+c-'0',c=getchar();    return ret*f;}#define space() putchar(' ')#define nextline() putchar('\n')void pot(int x){if(!x)return;pot(x/10);putchar('0'+x%10);}void out(int x){if(!x)putchar('0');if(x<0)putchar('-'),x=-x;pot(x);}  const int MAXN = 2005;const int MOD = 10; inline int mul(int x,int y){return (1ll*x*y)%MOD;}inline void Mul(int &x,int y){x=mul(x,y);}inline int sub(int x,int y){return x-y<0?x-y+MOD:x-y;}inline int add(int x,int y){if(y>=0)return x+y>MOD?x+y-MOD:x+y;else return sub(x,-y);}inline void Add(int &x,int y){x=add(x,y);}   int n,m;int v[MAXN];int f[MAXN],g[MAXN][MAXN]; int main(){    n=rd();m=rd();    for(int i=1;i<=n;i++)v[i]=rd();    f[0]=1;    for(int i=1;i<=n;i++){        for(int j=m;j>=v[i];j--){            (f[j]+=f[j-v[i]])%=MOD;         }    }    for(int i=1;i<=n;i++){        g[i][0]=1;        for(int j=1;j<=m;j++){            if(j