URAL - 1036 Lucky Tickets(DP+大数)

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++]
4000
= 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
[s]*dp
[s]);
cout<<ans<<endl;
}
return 0;
}