URAL

题目描述：

点击打开链接

题意：一个2*N位的数要保证前N位数的和与后N位相等，并且这2*N位的和为s，求这样的数有多少个，首先s为奇数的时候答案肯定为0，然后s为偶数的时候那就是一个N位数和为s/2的方案数，简单DP，dp[i][j]表示i位数和为j的方案数进行状态转移即可，题目数据很大要用大数。

AC代码：

#include<iostream>#include<time.h>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<stack>#include<queue>#include<vector>#include<set>#include<algorithm>using namespace std;const double PI=acos(-1);const int SIZE=110;const int INF=0x3f3f3f3f;const int MAXM=2000;struct BigInteger {    int len, s[SIZE + 5];    BigInteger () {        memset(s, 0, sizeof(s));        len = 1;    }    BigInteger operator = (const char *num) { //字符串赋值        memset(s, 0, sizeof(s));        len = strlen(num);        for(int i = 0; i < len; i++) s[i] = num[len - i - 1] - '0';        return *this;    }    BigInteger operator = (const int num) { //int 赋值        memset(s, 0, sizeof(s));        char ss[SIZE + 5];        sprintf(ss, "%d", num);        *this = ss;        return *this;    }    BigInteger (int num) {        *this = num;    }    BigInteger (char* num) {        *this = num;    }    string str() const { //转化成string        string res = "";        for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;        if(res == "") res = "0";        return res;    }    BigInteger clean() {        while(len > 1 && !s[len - 1]) len--;        return *this;    }    BigInteger operator + (const BigInteger& b) const {        BigInteger c;        c.len = 0;        for(int i = 0, g = 0; g || i < max(len, b.len); i++) {            int x = g;            if(i < len) x += s[i];            if(i < b.len) x += b.s[i];            c.s[c.len++] = x % 10;            g = x / 10;        }        return c.clean();    }    BigInteger operator - (const BigInteger& b) {        BigInteger c;        c.len = 0;        for(int i = 0, g = 0; i < len; i++) {            int x = s[i] - g;            if(i < b.len) x -= b.s[i];            if(x >= 0) g = 0;            else {                g = 1;                x += 10;            }            c.s[c.len++] = x;        }        return c.clean();    }    BigInteger operator * (const int num) const {        int c = 0, t;        BigInteger pro;        for(int i = 0; i < len; ++i) {            t = s[i] * num + c;            pro.s[i] = t % 10;            c = t / 10;        }        pro.len = len;        while(c != 0) {            pro.s[pro.len++] = c % 10;            c /= 10;        }        return pro.clean();    }    BigInteger operator * (const BigInteger& b) const {        BigInteger c;        for(int i = 0; i < len; i++) {            for(int j = 0; j < b.len; j++) {                c.s[i + j] += s[i] * b.s[j];                c.s[i + j + 1] += c.s[i + j] / 10;                c.s[i + j] %= 10;            }        }        c.len = len + b.len + 1;        return c.clean();    }    BigInteger operator / (const BigInteger &b) const {        BigInteger c, f;        for(int i = len - 1; i >= 0; --i) {            f = f * 10;            f.s[0] = s[i];            while(f >= b) {                f = f - b;                ++c.s[i];            }        }        c.len = len;        return c.clean();    }    //高精度取模    BigInteger operator % (const BigInteger &b) const{        BigInteger r;        for(int i = len - 1; i >= 0; --i) {            r = r * 10;            r.s[0] = s[i];            while(r >= b) r = r - b;        }        r.len = len;        return r.clean();    }    bool operator < (const BigInteger& b) const {        if(len != b.len) return len < b.len;        for(int i = len - 1; i >= 0; i--)            if(s[i] != b.s[i]) return s[i] < b.s[i];        return false;    }    bool operator > (const BigInteger& b) const {        return b < *this;    }    bool operator <= (const BigInteger& b) const {        return !(b < *this);    }    bool operator == (const BigInteger& b) const {        return !(b < *this) && !(*this < b);    }    bool operator != (const BigInteger &b) const {        return !(*this == b);    }    bool operator >= (const BigInteger &b) const {        return *this > b || *this == b;    }    friend istream & operator >> (istream &in, BigInteger& x) {        string s;        in >> s;        x = s.c_str();        return in;    }    friend ostream & operator << (ostream &out, const BigInteger& x) {        out << x.str();        return out;    }};int n,s;BigInteger dp[55][1010];;int main(){    while(scanf("%d%d",&n,&s)!=EOF)    {        if (s&1) {            printf("0\n");            continue;        }        s=s/2;        memset(dp,0,sizeof(dp));        for (int i=0;i<=9;i++)            dp[1][i]=1;        for (int i=2;i<=n;i++) {            for (int j=s;j>=0;j--) {                for (int k=0;k<=9;k++) {                    if (j-k>=0&&dp[i-1][j-k]!=0) {                        dp[i][j]=dp[i][j]+dp[i-1][j-k];                    }                }            }        }        BigInteger ans=(dp[n][s]*dp[n][s]);        cout<<ans<<endl;    }    return 0;}