BZOJ2179 FFT快速傅立叶

这题目，看了标题就知道要干啥了233

于是蒟蒻开始现学FFT。。。然后FFTing。。。

最后WA3+1AC。。。

WA了是怎么回事呢？

我先位压4位。。。爆long long了。。。然后位压3位，又爆long long了额T T

于是只好位压2位了233

 

  1 /**************************************************************
  2     Problem: 2179
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:616 ms
  7     Memory:24248 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13  
 14 #define complex P
 15 using namespace std;
 16 typedef double lf;
 17 typedef long long ll;
 18 const int N = 300005;
 19 const lf pi = acos(-1);
 20  
 21 struct complex {
 22     lf x, y;
 23     P() {}
 24     P(lf _x, lf _y) : x(_x), y(_y) {}
 25      
 26     P operator + (const P &b) {
 27         return P(x + b.x, y + b.y);
 28     }
 29     P operator - (const P &b) {
 30         return P(x - b.x, y - b.y);
 31     }
 32     P operator * (const P &b) {
 33         return P(x * b.x - y * b.y, x * b.y + y * b.x);
 34     }
 35 } x[N], y[N], w[2][N];
 36  
 37 int n, k;
 38 ll a[N], b[N];
 39  
 40 void read(ll *a) {
 41     int i, now = ceil((lf) n / 2) - 1;
 42     char ch = getchar();
 43     while (ch < '0' || '9' < ch) ch = getchar();
 44     for (i = n; i; --i) {
 45         (a[now] *= 10) += ch - '0';
 46         ch = getchar();
 47         if (i & 1) --now;
 48     }
 49 }
 50  
 51 void print(ll *a, int n) {
 52     printf("%lld", a[n--]);
 53     for (; ~n; --n)
 54         printf("%02lld", a[n]);
 55 }
 56  
 57  
 58 void FFT(P *x, int k, int v) {
 59     int i, j, l;
 60     P tmp;
 61     for (i = j = 0; i < k; ++i) {
 62         if (i > j) swap(x[i], x[j]);
 63         for (l = k >> 1; (j ^= l) < l; l >>= 1);
 64     }
 65     for (i = 2; i <= k; i <<= 1)
 66         for (j = 0; j < k; j += i)
 67             for (l = 0; l < i >> 1; ++l) {
 68                 tmp = x[j + l + (i >> 1)] * w[v][k / i * l];
 69                 x[j + l + (i >> 1)] = x[j + l] - tmp;
 70                 x[j + l] = x[j + l] + tmp;
 71             }
 72 }
 73  
 74 void work() {
 75     int i;
 76     for (k = 1; k < n << 1; k <<= 1);
 77     for (i = 0; i <= k; ++i)
 78         w[1][k - i] = w[0][i] = P(cos(pi * 2 * i / k), sin(pi * 2 * i / k));
 79     for (i = 0; i < k; ++i)
 80         x[i] = P(a[i], 0);
 81     FFT(x, k, 0);
 82     for (i = 0; i < k; ++i)
 83         y[i] = P(b[i], 0);
 84     FFT(y, k, 0);
 85     for (i = 0; i < k; ++i)
 86         x[i] = x[i] * y[i];
 87     FFT(x, k, 1);
 88     for (i = 0; i < 2 * k; ++i)
 89         a[i] = (int) (x[i].x / k + 0.5);
 90     for (i = 0; i < 2 * k - 1; ++i)
 91         a[i + 1] += a[i] / 100, a[i] %= 100;
 92     for (k = 2 * k - 1; !a[k]; --k);
 93     print(a, k);
 94 }
 95  
 96 int main() {
 97     int i;
 98     scanf("%d", &n);
 99     read(a), read(b);
100     n = ceil((lf) n / 2);
101     work();
102     return 0;   
103 }
View Code

（p.s. 感觉这种写法比算导的简单好多而且好理解的说= =）

posted on 2014-11-29 11:59  Xs酱~  阅读(193)  评论(0)  编辑  收藏  举报