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

/**************************************************************
    Problem: 2179
    User: rausen
    Language: C++
    Result: Accepted
    Time:616 ms
    Memory:24248 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <algorithm>
 
#define complex P
using namespace std;
typedef double lf;
typedef long long ll;
const int N = 300005;
const lf pi = acos(-1);
 
struct complex {
    lf x, y;
    P() {}
    P(lf _x, lf _y) : x(_x), y(_y) {}
     
    P operator + (const P &b) {
        return P(x + b.x, y + b.y);
    }
    P operator - (const P &b) {
        return P(x - b.x, y - b.y);
    }
    P operator * (const P &b) {
        return P(x * b.x - y * b.y, x * b.y + y * b.x);
    }
} x[N], y[N], w[2][N];
 
int n, k;
ll a[N], b[N];
 
void read(ll *a) {
    int i, now = ceil((lf) n / 2) - 1;
    char ch = getchar();
    while (ch < '0' || '9' < ch) ch = getchar();
    for (i = n; i; --i) {
        (a[now] *= 10) += ch - '0';
        ch = getchar();
        if (i & 1) --now;
    }
}
 
void print(ll *a, int n) {
    printf("%lld", a[n--]);
    for (; ~n; --n)
        printf("%02lld", a[n]);
}
 
 
void FFT(P *x, int k, int v) {
    int i, j, l;
    P tmp;
    for (i = j = 0; i < k; ++i) {
        if (i > j) swap(x[i], x[j]);
        for (l = k >> 1; (j ^= l) < l; l >>= 1);
    }
    for (i = 2; i <= k; i <<= 1)
        for (j = 0; j < k; j += i)
            for (l = 0; l < i >> 1; ++l) {
                tmp = x[j + l + (i >> 1)] * w[v][k / i * l];
                x[j + l + (i >> 1)] = x[j + l] - tmp;
                x[j + l] = x[j + l] + tmp;
            }
}
 
void work() {
    int i;
    for (k = 1; k < n << 1; k <<= 1);
    for (i = 0; i <= k; ++i)
        w[1][k - i] = w[0][i] = P(cos(pi * 2 * i / k), sin(pi * 2 * i / k));
    for (i = 0; i < k; ++i)
        x[i] = P(a[i], 0);
    FFT(x, k, 0);
    for (i = 0; i < k; ++i)
        y[i] = P(b[i], 0);
    FFT(y, k, 0);
    for (i = 0; i < k; ++i)
        x[i] = x[i] * y[i];
    FFT(x, k, 1);
    for (i = 0; i < 2 * k; ++i)
        a[i] = (int) (x[i].x / k + 0.5);
    for (i = 0; i < 2 * k - 1; ++i)
        a[i + 1] += a[i] / 100, a[i] %= 100;
    for (k = 2 * k - 1; !a[k]; --k);
    print(a, k);
}
 
int main() {
    int i;
    scanf("%d", &n);
    read(a), read(b);
    n = ceil((lf) n / 2);
    work();
    return 0;   
}

（p.s. 感觉这种写法比算导的简单好多而且好理解的说= =）