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.
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
Recommend
【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(" "); 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 }