方程的解

 

 sol：将问题转换为X1+X2+...+Xn=P-(A1+A2+...+An),X1,X2,...,Xn>=0。因此ans=C(n+P-(A1+A2+...+An)-1,n-1).

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())   x=(x<<3)+(x<<1)+ch-'0';
    return x*f;
}
inline void print(int x){
    if (x>=10) print(x/10);
    putchar(x%10+'0');
}
ll C(int n,int m){
    ll res=1;
    for (int i=1;i<=m;i++)   res=res*(n+1-i)/i;
    return res;
}
int main(){
    int n=read(),tot=0;
    for (int i=1;i<=n;i++)   tot+=read();
    int p=read()-tot;
    //既然每个数字都要有个下限，那就先从总和中减去就好了
    //接下来套公式求解
    printf("%lld
",C(n+p-1,n-1));
    return 0;
}