hdu 1223 Order Count

http://acm.hdu.edu.cn/showproblem.php?pid=1223

　　dp递推求满足条件的不等式的个数。

　　这里需要用到大数乘法，于是我就顺便打了一个大数的模板，重载了+和*运算符。

　　递推的公式很简单，dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j，其中i 是指共有i 个符号，j 是指将i 个符号分成j 份，dp得到种类的数目。然后dp[i][1~i]各项相加，就是最终的答案了。

代码如下：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cassert>
#include <cmath>
#include <algorithm>

using namespace std;

const int maxSize = 150;

struct bigNum {
    char num[maxSize];
};

bigNum operator * (bigNum a, bigNum b) {
    int la = strlen(a.num);
    int lb = strlen(b.num);
    bigNum ret;
    int tmp[maxSize];

    for (int i = 0; i < la; i++) a.num[i] -= '0';
    for (int i = 0; i < lb; i++) b.num[i] -= '0';
    for (int i = 0, endi = la + lb; i <= endi; i++) tmp[i] = ret.num[i] = 0;
    for (int i = 0; i < la; i++) {
        int ti = la - 1 - i;

        for (int j = 0; j < lb; j++) {
            int tj = lb - 1 - j;

            tmp[i + j] += a.num[ti] * b.num[tj];
        }
    }
    for (int i = 0, endi = la + lb - 1; i < endi; i++) {
        tmp[i + 1] += tmp[i] / 10;
        ret.num[endi - i] = tmp[i] % 10 + '0';
    }
    ret.num[0] = tmp[la + lb - 2] / 10 + '0';

    if (ret.num[0] == '0') {
        for (int i = 1, endi = la + lb + 2; i < endi; i++) {
            ret.num[i - 1] = ret.num[i];
        }
    }

    return ret;
}

bigNum operator + (bigNum a, bigNum b) {
    int la = strlen(a.num);
    int lb = strlen(b.num);
    bigNum ret;
    int maxLen = max(la, lb);
    int minLen = min(la, lb);

    for (int i = 0; i <= maxLen + 1; i++) ret.num[i] = 0;
    for (int i = 0; i < minLen; i++) {
        ret.num[maxLen - i] += a.num[la - i - 1] + b.num[lb - i - 1] - '0';
        if (ret.num[maxLen - i] > '9') ret.num[maxLen - i - 1]++, ret.num[maxLen - i] -= 10;
    }
    if (maxLen == la) {
        for (int i = minLen; i < maxLen; i++) {
            ret.num[maxLen - i] += a.num[la - i - 1];
            if (ret.num[maxLen - i] > '9') ret.num[maxLen - i - 1]++, ret.num[maxLen - i] -= 10;
        }
    } else {
        for (int i = minLen; i < maxLen; i++) {
            ret.num[maxLen - i] += b.num[lb - i - 1];
            if (ret.num[maxLen - i] > '9') ret.num[maxLen - i - 1]++, ret.num[maxLen - i] -= 10;
        }
    }
    ret.num[0] += '0';

    if (ret.num[0] == '0') {
        for (int i = 1, endi = la + lb + 2; i < endi; i++) {
            ret.num[i - 1] = ret.num[i];
        }
    }

    return ret;

}

const int maxn = 52;
bigNum dp[maxn][maxn];
bigNum fac[maxn];

char *con(int x) {
    int len = (int)log10((double)x) + 1;
    char *ret = new char[len + 1];

    for (int i = len - 1; i >= 0; i--) {
        ret[i] = x % 10 + '0';
        x /= 10;
    }
    ret[len] = 0;

    return ret;
}

void pre() {
    strcpy(fac[0].num, "1");
    for (int i = 1; i < maxn; i++){
        bigNum tmp;
        char *temp = con(i);

        strcpy(tmp.num, temp);
        delete temp;
        fac[i] = fac[i - 1] * tmp;
    }
    for (int i = 1; i < maxn; i++) {
        strcpy(dp[i][1].num, "1");
        strcpy(dp[i][i].num, "1");
        for (int j = 2; j < i; j++) {
            bigNum tmp;
            char *temp = con(j);

            strcpy(tmp.num, temp);
            delete temp;
            dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * tmp;
        }
        strcpy(dp[i][0].num, "0");
        for (int j = 1; j <= i; j++){
            dp[i][0] = dp[i][0] + dp[i][j] * fac[j];
        }
//printf("%s\n", dp[i][0].num);
    }
}

int main() {
    pre();
    int n, T;

    scanf("%d", &T);
    while (T-- && ~scanf("%d", &n)){
        printf("%s\n", dp[n][0].num);
    }

    return 0;
}

 

——written by Lyon