本文共 3774 字，大约阅读时间需要 12 分钟。
大数每次写都好麻烦这次转了一篇模板感觉不错哦
转：
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; const int MAXL = 500; struct BigNum { int num[MAXL]; int len; }; //高精度比较 a > b return 1, a == b return 0; a < b return -1; int Comp(BigNum &a, BigNum &b) { int i; if(a.len != b.len) return (a.len > b.len) ? 1 : -1; for(i = a.len-1; i >= 0; i--) if(a.num[i] != b.num[i]) return (a.num[i] > b.num[i]) ? 1 : -1; return 0; } //高精度加法 BigNum Add(BigNum &a, BigNum &b) { BigNum c; int i, len; len = (a.len > b.len) ? a.len : b.len; memset(c.num, 0, sizeof(c.num)); for(i = 0; i < len; i++) { c.num[i] += (a.num[i]+b.num[i]); if(c.num[i] >= 10) { c.num[i+1]++; c.num[i] -= 10; } } if(c.num[len]) len++; c.len = len; return c; } //高精度减法,保证a >= b BigNum Sub(BigNum &a, BigNum &b) { BigNum c; int i, len; len = (a.len > b.len) ? a.len : b.len; memset(c.num, 0, sizeof(c.num)); for(i = 0; i < len; i++) { c.num[i] += (a.num[i]-b.num[i]); if(c.num[i] < 0) { c.num[i] += 10; c.num[i+1]--; } } while(c.num[len] == 0 && len > 1) len--; c.len = len; return c; } //高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析 //如果b很大可以考虑把b看成高精度 BigNum Mul1(BigNum &a, int &b) { BigNum c; int i, len; len = a.len; memset(c.num, 0, sizeof(c.num)); //乘以0，直接返回0 if(b == 0) { c.len = 1; return c; } for(i = 0; i < len; i++) { c.num[i] += (a.num[i]*b); if(c.num[i] >= 10) { c.num[i+1] = c.num[i]/10; c.num[i] %= 10; } } while(c.num[len] > 0) { c.num[len+1] = c.num[len]/10; c.num[len++] %= 10; } c.len = len; return c; } //高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度， //如果确定不会发生溢出, 可以将里面的while改成if BigNum Mul2(BigNum &a, BigNum &b) { int i, j, len = 0; BigNum c; memset(c.num, 0, sizeof(c.num)); for(i = 0; i < a.len; i++) { for(j = 0; j < b.len; j++) { c.num[i+j] += (a.num[i]*b.num[j]); if(c.num[i+j] >= 10) { c.num[i+j+1] += c.num[i+j]/10; c.num[i+j] %= 10; } } } len = a.len+b.len-1; while(c.num[len-1] == 0 && len > 1) len--; if(c.num[len]) len++; c.len = len; return c; } //高精度除以低精度,除的结果为c, 余数为f void Div1(BigNum &a, int &b, BigNum &c, int &f) { int i, len = a.len; memset(c.num, 0, sizeof(c.num)); f = 0; for(i = a.len-1; i >= 0; i--) { f = f*10+a.num[i]; c.num[i] = f/b; f %= b; } while(len > 1 && c.num[len-1] == 0) len--; c.len = len; } //高精度*10 void Mul10(BigNum &a) { int i, len = a.len; for(i = len; i >= 1; i--) a.num[i] = a.num[i-1]; a.num[i] = 0; len++; //if a == 0 while(len > 1 && a.num[len-1] == 0) len--; } //高精度除以高精度，除的结果为c，余数为f void Div2(BigNum &a, BigNum &b, BigNum &c, BigNum &f) { int i, len = a.len; memset(c.num, 0, sizeof(c.num)); memset(f.num, 0, sizeof(f.num)); f.len = 1; for(i = len-1;i >= 0;i--) { Mul10(f); //余数每次乘10 f.num[0] = a.num[i]; //然后余数加上下一位 ///利用减法替换除法 while(Comp(f, b) >= 0) { f = Sub(f, b); c.num[i]++; } } while(len > 1 && c.num[len-1] == 0) len--; c.len = len; } void print(BigNum &a) //输出大数 { int i; for(i = a.len-1; i >= 0; i--) printf("%d", a.num[i]); puts(""); } //将字符串转为大数存在BigNum结构体里面 BigNum ToNum(char *s) { int i, j; BigNum a; a.len = strlen(s); for(i = 0, j = a.len-1; s[i] != '\0'; i++, j--) a.num[i] = s[j]-'0'; return a; } void Init(BigNum &a, char *s, int &tag) //将字符串转化为大数{ int i = 0, j = strlen(s); if(s[0] == '-') { j--; i++; tag *= -1; } a.len = j; for(; s[i] != '\0'; i++, j--) a.num[j-1] = s[i]-'0';} int main(void) { BigNum a, b; char s1[100], s2[100]; while(scanf("%s %s", s1, s2) != EOF) { int tag = 1; Init(a, s1, tag); //将字符串转化为大数 Init(b, s2, tag); a = Mul2(a, b); if(a.len == 1 && a.num[0] == 0) { puts("0"); } else { if(tag < 0) putchar('-'); print(a); } } return 0; }
转载于:https://www.cnblogs.com/Arthas0v0/p/3459477.html