方程的解

Posted 2021-03-21 cutepota

 

 

 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).

 

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<algorithm>
 6 #define inf 0x7f7f7f
 7 typedef long long ll;
 8 typedef unsigned int ui;
 9 typedef unsigned long long ull;
10 using namespace std;
11 inline int read(){
12     int x=0,f=1;char ch=getchar();
13     for (;ch<‘0‘||ch>‘9‘;ch=getchar())  if (ch==‘-‘)    f=-1;
14     for (;ch>=‘0‘&&ch<=‘9‘;ch=getchar())   x=(x<<3)+(x<<1)+ch-‘0‘;
15     return x*f;
16 }
17 inline void print(int x){
18     if (x>=10) print(x/10);
19     putchar(x%10+‘0‘);
20 }
21 ll C(int n,int m){
22     ll res=1;
23     for (int i=1;i<=m;i++)   res=res*(n+1-i)/i;
24     return res;
25 }
26 int main(){
27     int n=read(),tot=0;
28     for (int i=1;i<=n;i++)   tot+=read();
29     int p=read()-tot;
30     //既然每个数字都要有个下限，那就先从总和中减去就好了
31     //接下来套公式求解
32     printf("%lld
",C(n+p-1,n-1));
33     return 0;
34 }