初中一題多解的例題，UVA10523 Very Easy !!!題解

代碼來源：DeathYmz

AC的C++語言程序如下：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=10010;
const int base=10;
struct inter{int num[maxn];int cnt; 
};
inter a,b;
//大數模擬加減法 ---模板可套用 
//
void muti(int k)// 
{for(int i=0;i<a.cnt;i++){a.num[i]*=k;}int i;for(i=0;i<a.cnt;i++){a.num[i+1]+=a.num[i]/base;a.num[i]%=base;}//剩余之后 while(a.num[i]){a.num[i+1]+=a.num[i]/base;a.num[i]%=base;i++;} while(i>=0&&!a.num[i]) i--;//去零 a.cnt=i+1; 
}
inter add(inter &x,inter &y)//相加 
{inter tmp;memset(tmp.num,0,sizeof(tmp.num));tmp.cnt=0;for(int i=0, j=0, g=0; ; i++, j++)  {if(g==0 && i>=x.cnt && j>=y.cnt) break;  int t = g;  if(i<x.cnt) t+= a.num[i];  if(j<y.cnt) t+= b.num[i];  //這個挺好tmp.num[tmp.cnt++] = t%base;  g = t/base;  } return tmp; 
} 
int main()
{int N,A;while(scanf("%d%d",&N,&A)!=EOF){if(A==0){printf("0\n");continue;}memset(a.num,0,sizeof(a.num)); memset(b.num,0,sizeof(b.num));b.cnt=a.cnt=0;for(int i=1;i<=N;i++){memset(a.num,0,sizeof(a.num));a.cnt=1,a.num[0]=1;for(int j=1;j<=i;j++)muti(A);//累乘 muti(i);b=add(a,b);}for(int i=b.cnt-1;i>=0;i--)printf("%d",b.num[i]);printf("\n");}return 0;
}