【HDU1402】【FNT版】A * B Problem Plus

Problem Description

Calculate A * B.

 

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

 

Output

For each case, output A * B in one line.

Sample Input

1
2
1000
2

Sample Output

2
2000

Author

DOOM III

 

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【分析】

写个数论变换还被卡出翔....

什么世道啊。。。。

 /*
 宋代朱敦儒
 《西江月·世事短如春梦》
 世事短如春梦，人情薄似秋云。不须计较苦劳心。万事原来有命。
 幸遇三杯酒好，况逢一朵花新。片时欢笑且相亲。明日阴晴未定。
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <iostream>
 #include <string>
 #include <ctime>
 #define LOCAL
 const int MAXN =  + ;
 const long long MOD = (479ll<<21ll) + ;//费马数论变换的费马素数
 long long G = ;//元根
 using namespace std;
 typedef long long ll;
 ll x1[MAXN], x2[MAXN];
 ll Inv[MAXN], wn[MAXN];
 ll Inv2[MAXN];
 char a[MAXN], b[MAXN];

 ll pow(ll a, ll b){
    if (b == ) return  % MOD;
    if (b == ) return a % MOD;
    ll tmp = pow(a, b / );
    if (b %  == ) return (tmp * tmp) % MOD;
    else return ((tmp * tmp) % MOD * (a % MOD)) % MOD;
 }
 ll exgcd(ll a, ll b, ll &x, ll &y){
    if (b == ){x = 1ll; y = ; return a;}
    ll tmp = exgcd(b, a % b, y, x);
    y -= x * (a / b);
    return tmp;
 }
 ll inv(ll a, ll p){
    ll x, y;
    ll tmp = exgcd(a, p, x, y);
    return ((x % MOD) + MOD) % MOD;
 }
 void change(ll *x, int len, int loglen){
      for (int i = ; i < len; i++){
          int t = i, k = , tmp = loglen;
          while (tmp--) {k = (k << ) + (t & ); t >>= ;}
          if (k < i) swap(x[i], x[k]);
      }
      return;
 }
 void FNT(ll *x, int len, int loglen, int type){
      if (type) change(x, len, loglen);
      int t;
      t = (type ? : ( << loglen));
      for (int i = ; i < loglen; i++){
          if (!type) t >>= ;
          int l = , r = l + t;
          while (l < len){
                ll a, b;
                ll tmp = 1ll, w = wn[t] % MOD;
                if (!type) w = Inv[t];
                for (int j = l; j < l + t; j++){
                    if (type){
                       a = x[j] % MOD;
                       b = (x[j + t] * (tmp % MOD)) % MOD;
                       x[j] = (a + b) % MOD;
                       x[j + t] = ((a - b) % MOD + MOD) % MOD;
                    }else{
                       a = (x[j] + x[j + t]) % MOD;
                       b = ((((x[j] - x[j + t]) % MOD + MOD) % MOD) * tmp) % MOD;
                       x[j] = a;
                       x[j + t] = b;
                    }
                    tmp = (tmp * w) % MOD;
                }
                l = r + t;
                r = l + t;
          }
          if (type) t <<= ;
      }
      if (!type){
         change(x, len, loglen);
         for (int i = ; i < len; i++) x[i] = (x[i] % MOD * Inv2[len]) % MOD;
      }
 }
 int work(char *a, char *b){
      int len1, len2, loglen = ;
      len1 = strlen(a);
      len2 = strlen(b);
      while (( << loglen) < (max(len1, len2) << )) loglen++;
      for (int i = ; i < len1; i++) x1[i] = 1ll * (a[i] - '');
      for (int i = ; i < len2; i++) x2[i] = 1ll * (b[i] - '');
      for (int i = len1; i < (<<loglen); i++) x1[i] = ;
      for (int i = len2; i < (<<loglen); i++) x2[i] = ;
      FNT(x1, (<<loglen), loglen, );
      FNT(x2, (<<loglen), loglen, );
      for (int i = ; i < (<<loglen); i++) x1[i] = (x1[i] * x2[i]) % MOD;
      FNT(x1, (<<loglen), loglen, );

      int len;
      ll over, t;
      over = ;
      len = ;
      for (int i = (len1 + len2) - ; i >= ; i--){
          t = x1[i] + over;
          a[len++] = t % ;
          over = t / ;
      }
      while (over){
            a[len++] = over % ;
            over /= ;
      }
      return len;
 }
 void print(char *t, int len){
      for (; len >=  && !t[len]; len--);
      if (len < ) printf("");
      else for (int i = len; i >= ; i--) printf("%c", t[i] + '');
      printf("\n");
 }
 void prepare(){
      wn[] = MOD - ;
      for (int i = ; i < ; i++){
          if (i !=  && (MOD - ) / (i * ) == (MOD - ) / ((i - ) * )){
             wn[i] = wn[i - ];
             Inv[i] = Inv[i - ];
          }else{
             wn[i] = pow(G, (MOD - ) / ( * i));
             Inv[i] = inv(wn[i], MOD);
          }
          Inv2[i] = inv(i, MOD);
      }
 }

 int main() {
     memset(a, , sizeof(a));
     memset(b, , sizeof(b));
     prepare();
     while (scanf("%s%s", a, b) != EOF){
           print(a, work(a, b));
           memset(a, , sizeof(a));
           memset(b, , sizeof(b));
     }
     return ;
 }