自己一点一点敲出来的，累！！
#include#include #include #define REP(i, a, b) for(int i = (a); i < (b); i++)#define _for(i, a, b) for(int i = (a); i <= (b); i++)using namespace std;const int MAXN = 112;const int base = 10000; //压4位效率高且省空间 struct node{ int len, s[505]; //s的范围开太大空间会炸，时间也会多 node() { len = 0; memset(s, 0, sizeof(s)); } //如果要表示0就直接 node ans就好了 }; //len = 0时表示值为0//比较大小bool judge(node a, node b) //这种写法很简略 { if(a.len != b.len) return a.len > b.len; for(int i = a.len; i >= 1; i--) if(a.s[i] != b.s[i]) return a.s[i] > b.s[i]; return true;}//高精度加法 node operator + (const node& a, const node& b) { node c; int& len = c.len = max(a.len, b.len); _for(i, 1, len) { c.s[i] += a.s[i] + b.s[i]; //这里一定是+=，不是= c.s[i+1] += c.s[i] / base; c.s[i] %= base; } if(c.s[len+1]) len++; return c;}//高精度减法 node operator - (bignum a, const bignum& b){ for(int i = a.len; i >= 1; i--) { a.s[i] -= b.s[i]; if(a.s[i] < 0) a.s[i+1]--, a.s[i] += base; } while(!a.s[a.len] && a.len > 0) a.len--; return a;}//高精度*高精度 node operator * (const node& a, const node& b){ node c; int& len = c.len = b.len + a.len - 1; _for(i, 1, a.len) _for(j, 1, b.len) { c.s[i+j-1] += a.s[i] * b.s[j]; c.s[i+j] += c.s[i+j-1] / base; c.s[i+j-1] %= base; } if(c.s[len+1]) len++; return c;}//低精度*高精度 node operator * (const int& a, const node& b) //系统会根据数据类型来判断是低精度乘高精度还是高精度乘高精度 { node c; int& len = c.len = b.len; _for(i, 1, b.len) { c.s[i] += b.s[i] * a; c.s[i+1] += c.s[i] / base; c.s[i] %= base; } while(c.s[len+1] > 0) { c.len++; c.s[len+1] += c.s[len] / base; c.s[len] %= base; } return c;}//读入char str[10000 + 5];void read(node& a) { scanf("%s", str); reverse(str, str + strlen(str)); //先翻转再说 int& len = a.len = 0; for(int i = 0, w; i < strlen(str); i++, w *= 10) { if(i % 4 == 0) len++, w = 1; a.s[len] += w * (str[i] - '0'); }}//输出void print(bignum a){ printf("%d", a.s[a.len]); for(int i = a.len - 1; i >= 1; i--) printf("%04d", a.s[i]); puts("");}