大整数开根 codevs3119

Description

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给出一个正整数n，求n开根号后的整数部分的值。n的位数不超过1000位

Solution

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我果然还是好弱啊

二分一个答案然后高精度乘法判断，然后我压位都打错一个晚上就过去了。。
这次的程序应该能当高精度的标了吧

Code

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#include <stdio.h>#include <string.h>#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)#define drp(i, st, ed) for (int i = st; i >= ed; i -= 1)#define fill(x, t) memset(x, t, sizeof(x))#define max(x, y) ((x)>(y)?(x):(y))#define L 1001#define MOD 1000struct num{    int s[L], len;    inline bool operator ==(num b){        num a = *this;        if (a.len != b.len){            return 0;        }else{            drp(i, a.len, 1){                if (a.s[i] != b.s[i]){                    return 0;                }            }            return 1;        }    }    inline bool operator <(num b){        num a = *this;        if (a.len < b.len){            return 1;        }else if (a.len > b.len){            return 0;        }else{            drp(i, a.len, 1){                if (a.s[i] < b.s[i]){                    return 1;                }else if (a.s[i] > b.s[i]){                    return 0;                }            }            return 0;        }    }    inline bool operator <=(num b){        num a = *this;        if (a < b || a == b){            return 1;        }else{            return 0;        }    }    inline bool operator >(num b){        num a = *this;        if (a.len > b.len){            return 1;        }else if (a.len < b.len){            return 0;        }else{            drp(i, a.len, 1){                if (a.s[i] > b.s[i]){                    return 1;                }else if (a.s[i] < b.s[i]){                    return 0;                }            }            return 0;        }    }    inline bool operator >=(num b){        num a = *this;        if (a > b || a == b){            return 1;        }else{            return 0;        }    }    inline num operator +(num b){        num a = *this, c = (num){{0}, max(a.len, b.len)};        int v = 0;        rep(i, 1, c.len){            c.s[i] = (a.s[i] + b.s[i] + v) % MOD;            v = (a.s[i] + b.s[i] + v) / MOD;        }        if (v){            c.len += 1;            c.s[c.len] = v;        }        return c;    }    inline num operator -(num b){        num a = *this, c = (num){{0}, max(a.len, b.len)};        rep(i, 1, c.len){            c.s[i] = a.s[i] - b.s[i];            if (c.s[i] < 0){                c.s[i] += MOD;                a.s[i + 1] -= 1;            }        }        while (!c.s[c.len] && c.len > 1){            c.len -= 1;        }        return c;    }    inline num operator *(num b){        num a = *this, c = (num){{0}, a.len + b.len};        rep(i, 1, a.len){            rep(j, 1, b.len){                c.s[i + j - 1] += a.s[i] * b.s[j];            }        }        rep(i, 1, a.len + b.len){            c.s[i + 1] += c.s[i] / MOD;            c.s[i] %= MOD;        }        while (!c.s[c.len] && c.len > 1){            c.len -= 1;        }        return c;    }    inline num operator /(int b){        num a = *this, c = (num){{0}, a.len};        int v = 0;        drp(i, len, 1){            int t = v * MOD + a.s[i];            c.s[i] = t / b;            v = t % b;        }        while (!c.s[c.len] && c.len > 1){            c.len -= 1;        }        return c;    }    inline void read(){        fill(s, 0); len = 0;        char st[L];        scanf("%s", st);        int v = 0, i;        for (i = strlen(st) - 1; i >= 3; i -= 3){            rep(j, i - 2, i){                v = v * 10 + st[j] - '0';            }            s[++ len] = v;            v = 0;        }        rep(j, 0, i){            v = v * 10 + st[j] - '0';        }        s[++ len] = v;    }    inline void read1(int x){        fill(s, 0);        len = 0;        do{s[++ len] = x % MOD;}while (x /= MOD);    }    inline void output(){        num tmp = *this;        int i = tmp.len;        while (!tmp.s[i] && i > 1){            i -= 1;        }        printf("%d", tmp.s[i]);        drp(j, i - 1, 1){            int v = tmp.s[j];            int f[5] = {0};            rep(k, 1, 3){                f[k] = v % 10;                v /= 10;            }            drp(k, 3, 1){                printf("%d", f[k]);            }        }        printf("\n");    }};num one;int main(void){    one.s[1] = 1;    one.len = 1;    num p;    p.read();    num r = p;    num l; l.s[1] = 0; l.len = 1;    num mid, ans;    while (l <= r){        mid = l + r;        mid = mid / 2;        num qr = mid * mid;        if (qr <= p){            l = mid + one;            ans = mid;        }else if (qr > p){            r = mid - one;        }    }    ans.output();    return 0;}