题目描述
佳佳碰到了一个难题，请你来帮忙解决。
对于不定方程a1+a2+…+ak-1+ak=g(x)，其中k≥2且k∈N，x是正整数，g(x)=x^x mod 1000(即x^x除以1000的余数)，x，k是给定的数。我们要求的是这个不定方程的正整数解组数。
举例来说，当k=3，x=2时，分别为(a1,a2,a3)=(2,1,1)'(1,2,1)，(1,1,2)。
输入输出格式
输入格式：输入文件equation.in有且只有一行，为用空格隔开的两个正整数，依次为k，x。
输出格式：输出文件equation.out有且只有一行，为方程的正整数解组数。
输入输出样例
说明
对于40%的数据，ans≤10^16；对于100%的数据，k≤100，x≤2^31-1，k≤g(x)。
_NOI导刊2010提高（01）
隔板法：
C(x-1,k-1) ；
然后高精度就行了；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 900005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
const int W = 10000000;
int x, k;
struct bigint {
int a[25], len;
bigint() {
ms(a); len = 0;
}
bigint operator +(const bigint &rhs)const {
bigint c; int x = 0;
c.len = max(len, rhs.len);
for (int i = 1; i <= c.len; i++) {
c.a[i] = a[i] + rhs.a[i] + x;
x = c.a[i] / W; c.a[i] %= W;
}
for (; x; x /= W)c.a[++c.len] = x % W;
return c;
}
void print() {
cout << a[len];
for (int i = len - 1; i >= 1; i--) {
for (int j = 10; a[i] * j < W; j *= 10)
putchar(48);
cout << a[i];
}
}
}c[1003][1003];
int qpow(int x, int y) {
int res = 1;
while (y) {
if (y % 2)res = 1ll * res*x % 1000;
x = x * x % 1000; y >>= 1;
}
return res;
}
int main() {
//ios::sync_with_stdio(0);
rdint(k); rdint(x); x %= 1000; x = qpow(x, x);
for (int i = 0; i < x; i++) {
for (int j = 0; j <= i; j++) {
if (!j || j == i)c[i][j].a[c[i][j].len = 1] = 1;
else c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}
}
c[x - 1][k - 1].print();
return 0;
}