【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。。。。

  1 /*

  2 宋代朱敦儒

  3 《西江月·世事短如春梦》

  4 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。

  5 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。 

  6 */

  7 #include <cstdio>

  8 #include <cstring>

  9 #include <algorithm>

 10 #include <cmath>

 11 #include <queue>

 12 #include <vector>

 13 #include <iostream>

 14 #include <string>

 15 #include <ctime>

 16 #define LOCAL

 17 const int MAXN = 202000 + 10;

 18 const long long MOD = (479ll<<21ll) + 1;//费马数论变换的费马素数 

 19 long long G = 3;//元根 

 20 using namespace std;

 21 typedef long long ll;

 22 ll x1[MAXN], x2[MAXN];

 23 ll Inv[MAXN], wn[MAXN];

 24 ll Inv2[MAXN];

 25 char a[MAXN], b[MAXN];

 26 

 27 ll pow(ll a, ll b){

 28    if (b == 0) return 1 % MOD;

 29    if (b == 1) return a % MOD;

 30    ll tmp = pow(a, b / 2);

 31    if (b % 2 == 0) return (tmp * tmp) % MOD;

 32    else return ((tmp * tmp) % MOD * (a % MOD)) % MOD;

 33 }

 34 ll exgcd(ll a, ll b, ll &x, ll &y){

 35    if (b == 0){x = 1ll; y = 0; return a;}

 36    ll tmp = exgcd(b, a % b, y, x);

 37    y -= x * (a / b);

 38    return tmp;

 39 }

 40 ll inv(ll a, ll p){

 41    ll x, y;

 42    ll tmp = exgcd(a, p, x, y);

 43    return ((x % MOD) + MOD) % MOD; 

 44 }

 45 void change(ll *x, int len, int loglen){

 46      for (int i = 0; i < len; i++){

 47          int t = i, k = 0, tmp = loglen;

 48          while (tmp--) {k = (k << 1) + (t & 1); t >>= 1;}

 49          if (k < i) swap(x[i], x[k]);

 50      }

 51      return;

 52 }

 53 void FNT(ll *x, int len, int loglen, int type){

 54      if (type) change(x, len, loglen);

 55      int t;

 56      t = (type ? 1: (1 << loglen));

 57      for (int i = 0; i < loglen; i++){

 58          if (!type) t >>= 1;

 59          int l = 0, r = l + t;

 60          while (l < len){

 61                ll a, b;

 62                ll tmp = 1ll, w = wn[t] % MOD;

 63                if (!type) w = Inv[t];

 64                for (int j = l; j < l + t; j++){

 65                    if (type){

 66                       a = x[j] % MOD;

 67                       b = (x[j + t] * (tmp % MOD)) % MOD;

 68                       x[j] = (a + b) % MOD;

 69                       x[j + t] = ((a - b) % MOD + MOD) % MOD; 

 70                    }else{

 71                       a = (x[j] + x[j + t]) % MOD;

 72                       b = ((((x[j] - x[j + t]) % MOD + MOD) % MOD) * tmp) % MOD;

 73                       x[j] = a;

 74                       x[j + t] = b;

 75                    }

 76                    tmp = (tmp * w) % MOD;

 77                }

 78                l = r + t;

 79                r = l + t;

 80          } 

 81          if (type) t <<= 1;

 82      }

 83      if (!type){

 84         change(x, len, loglen);

 85         for (int i = 0; i < len; i++) x[i] = (x[i] % MOD * Inv2[len]) % MOD;

 86      }

 87 }

 88 int work(char *a, char *b){

 89      int len1, len2, loglen = 0;

 90      len1 = strlen(a);

 91      len2 = strlen(b);

 92      while ((1 << loglen) < (max(len1, len2) << 1)) loglen++;

 93      for (int i = 0; i < len1; i++) x1[i] = 1ll * (a[i] - '0');

 94      for (int i = 0; i < len2; i++) x2[i] = 1ll * (b[i] - '0');

 95      for (int i = len1; i < (1<<loglen); i++) x1[i] = 0;

 96      for (int i = len2; i < (1<<loglen); i++) x2[i] = 0;

 97      FNT(x1, (1<<loglen), loglen, 1);

 98      FNT(x2, (1<<loglen), loglen, 1);

 99      for (int i = 0; i < (1<<loglen); i++) x1[i] = (x1[i] * x2[i]) % MOD;

100      FNT(x1, (1<<loglen), loglen, 0);

101      

102      int len;

103      ll over, t;

104      over = 0;

105      len = 0;

106      for (int i = (len1 + len2) - 2; i >= 0; i--){

107          t = x1[i] + over;

108          a[len++] = t % 10;

109          over = t / 10;

110      }

111      while (over){

112            a[len++] = over % 10;

113            over /= 10;

114      }

115      return len;

116 }

117 void print(char *t, int len){

118      for (; len >= 0 && !t[len]; len--);

119      if (len < 0) printf("0");

120      else for (int i = len; i >= 0; i--) printf("%c", t[i] + '0');

121      printf("\n");

122 }

123 void prepare(){

124      wn[0] = MOD - 1;

125      for (int i = 1; i < 200010; i++){

126          if (i != 1 && (MOD - 1) / (i * 2) == (MOD - 1) / ((i - 1) * 2)){

127             wn[i] = wn[i - 1];

128             Inv[i] = Inv[i - 1];

129          }else{

130             wn[i] = pow(G, (MOD - 1) / (2 * i));

131             Inv[i] = inv(wn[i], MOD);

132          }

133          Inv2[i] = inv(i, MOD); 

134      }

135 }

136 

137 int main() {

138     memset(a, 0, sizeof(a));

139     memset(b, 0, sizeof(b));

140     prepare();

141     while (scanf("%s%s", a, b) != EOF){

142           print(a, work(a, b));

143           memset(a, 0, sizeof(a));

144           memset(b, 0, sizeof(b));

145     }

146     return 0;

147 }

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