toj1746How Many Sums
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这题题目大意是给你一个数,问把这个数拆成若干个数相加,可以有多少种拆法.开始没甚么思路,后来想到用dp做.定义状态dp[i][j]为把数字i拆成j个数相加有多少种拆法.其实有了思路就简单了:拆数字不如合并数字来得容易,把某个数n拆成x(1<=x<=n)份,等价于把n个1随意合并,合并成1到n份.然后看下面:从例子出发,比如10这个数,我试着拆分,可以分成1~10份.1.分5~10份时(也就是把n拆成>=n/2份),其实就相当于从10个1中拿出0~5个,然后把它们搞一搞,可以两两组合等等,反正不管怎么搞,既然是动态规划,我们根本不需要在意此等细节,具体来说,分5份时,就是留5个1不动,把其它5个1搞成1到5份,其实就相当于求解把5分成若干个数相加,有多少种分法,这时我们看到了子问题的身影了,所以等价于dp[5][k] k=1,2,...5分6份,时,就是留6个1不动,拿出剩下的4个1搞一搞,搞成1~4份然后丢向这6个1,由于都是1,所以它们之间是等价的.有dp[4][k],k=1,2,3,4种分法分7份时,就是留7个1不动,拿出剩下的3个1搞成1~3份再砸那7个1,砸到谁就和谁结合了...dp[3][1]+dp[3][2]+dp[3][3]如此一来,我们可以知道dp[10][j](5<=j<=10) = sum{dp[i-j][k]} (k=1,2,...i-j)2.分1~4份时(也就是把n拆成<=n/2-1份,小于一半),就不能和上面那么搞了,因为这时比如我拿出6个1,搞一搞,不能搞成5份然后和剩下的4个1结合,因为只剩4个不动的1了~~但是我们可以知道,这6个1可以分成1份,2份,3份,最多分4份,所以dp[10][4] = dp[6][k](k=1,2,3,4)其它也是类似的,然后可以得出dp[10][j](1<=j<=4) = sum{dp[i-j][k]} (k=1,2,...j)当然,上面两个式子可以合并:dp[10][j](1<=j<=10) = sum{dp[i-j][k]} (1<=k<=max{j, i-j})3.经过上面的例子分析,得到最终的状态转移方程:dp[i][j] = sum{dp[i-j][k]}, 1<=k<=max{j, i-j}, 1<=j<=i.(j是份数,所以当然是属于[1,n]的)4.但是故事还没有结束,这题最坑爹的地方在于它的整数可以达到600,从上面累加的夸张程度就可以看出来,这尼玛比斐波那契数列涨得还快!挂了两次,痛定思痛,我是后来才发现这尼玛随便一个第200项就爆long long 了!!!所以找了一个大整数的模板,然后乖乖AC...大整数部分的模板可以不用在意,直接拿来用就好了,重点还是后面的dp计算.另外我定义了dp[i][0]为把数字i分成若干份的结果也就是dp[i][0] = dp[i][1]+dp[i][2]+...+dp[i][i-1]+dp[i][i],这样在把n分成大于n/2份时,要用到的sum{dp[i-j][k]}, k=1,2,...,i-j就可以直接取用dp[i-j][0]了,因为dp[i-j][0]就等于dp[i-j][1]+dp[i-j][2]+...+dp[i-j][i-j].#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <bitset>#include <string>#include <cstring>#include <climits>#include <cstdio>#include <iostream>#include <algorithm>using namespace std;class BigNum;istream& operator>>(istream&, BigNum&);ostream& operator<<(ostream&, BigNum&);#define MAXN 9999#define MAXSIZE 40#define DLEN 4class BigNum {public: int a[MAXSIZE]; int len;public: BigNum(){len = 1;memset(a,0,sizeof(a));} BigNum(const int); BigNum(const char*); BigNum(const BigNum &); BigNum &operator=(const BigNum &); friend istream& operator>>(istream&, BigNum&); friend ostream& operator<<(ostream&, BigNum&); BigNum operator+(const BigNum &) const; BigNum operator-(const BigNum &) const; BigNum operator*(const BigNum &) const; BigNum operator/(const int &) const; BigNum operator^(const int &) const; int operator%(const int &) const; bool operator>(const BigNum & T)const; bool operator==(const BigNum & T)const; bool operator==(const int & t)const; bool operator>(const int & t)const;};istream& operator>>(istream& in, BigNum& b) { char ch[MAXSIZE*4]; int i = -1; in >> ch; int l = strlen(ch); int cnt = 0, sum = 0; for (i=l-1;i>=0;) { sum = 0; int t = 1; for (int j = 0;j < 4 && i >= 0;j++, i--, t*=10) { sum+=(ch[i]-'0')*t; } b.a[cnt]=sum; cnt++; } b.len = cnt++; return in;}ostream& operator<<(ostream& out, BigNum& b) { int i; cout << b.a[b.len - 1]; for (i = b.len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << b.a[i]; } return out;}BigNum::BigNum(const int b){ int c,d = b; len = 0; memset(a,0,sizeof(a)); while (d > MAXN) { c = d - (d / (MAXN + 1)) * (MAXN + 1); d = d / (MAXN + 1); a[len++] = c; } a[len++] = d;}BigNum::BigNum(const char*s){ int t,k,index,l; memset(a,0,sizeof(a)); l=strlen(s); len=l/DLEN; if (l%DLEN)len++; index=0; for (int i=l-1;i>=0;i-=DLEN) { t=0;k=i-DLEN+1; if (k<0)k=0; for (int j=k;j<=i;j++) t=t*10+s[j]-'0'; a[index++]=t; }}BigNum::BigNum(const BigNum & T) : len(T.len){ int i; memset(a,0,sizeof(a)); for (i = 0 ; i < len ; i++) a[i] = T.a[i];}BigNum & BigNum::operator=(const BigNum & n){ len = n.len; memset(a,0,sizeof(a)); for (int i = 0 ; i < len ; i++) a[i] = n.a[i]; return *this;}BigNum BigNum::operator+(const BigNum & T) const{ BigNum t(*this); int i,big; big = T.len > len ? T.len : len; for (i = 0 ; i < big ; i++) { t.a[i] +=T.a[i]; if (t.a[i] > MAXN) { t.a[i + 1]++; t.a[i] -=MAXN+1; } } if (t.a[big] != 0) t.len = big + 1; else t.len = big; return t;}BigNum BigNum::operator-(const BigNum & T) const{ int i,j,big; bool flag; BigNum t1,t2; if (*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for (i = 0 ; i < big ; i++) { if (t1.a[i] < t2.a[i]) { j = i + 1; while (t1.a[j] == 0) j++; t1.a[j--]--; while (j > i) t1.a[j--] += MAXN; t1.a[i] += MAXN + 1 - t2.a[i]; } else t1.a[i] -= t2.a[i]; } t1.len = big; while (t1.a[len - 1] == 0 && t1.len > 1) { t1.len--; big--; } if (flag)t1.a[big-1]=0-t1.a[big-1]; return t1;}BigNum BigNum::operator*(const BigNum & T) const{ BigNum ret; int i,j,up; int temp,temp1; for (i = 0 ; i < len ; i++) { up = 0; for (j = 0 ; j < T.len ; j++) { temp = a[i] * T.a[j] + ret.a[i + j] + up; if (temp > MAXN) { temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); up = temp / (MAXN + 1); ret.a[i + j] = temp1; } else { up = 0; ret.a[i + j] = temp; } } if (up != 0) ret.a[i + j] = up; } ret.len = i + j; while (ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret;}BigNum BigNum::operator/(const int & b) const{ BigNum ret; int i,down = 0; for (i = len - 1 ; i >= 0 ; i--) { ret.a[i] = (a[i] + down * (MAXN + 1)) / b; down = a[i] + down * (MAXN + 1) - ret.a[i] * b; } ret.len = len; while (ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret;}int BigNum::operator %(const int & b) const{ int i,d=0; for (i = len-1; i>=0; i--) { d = ((d * (MAXN+1))% b + a[i])% b; } return d;}BigNum BigNum::operator^(const int & n) const{ BigNum t,ret(1); int i; if (n<0)exit(-1); if (n==0)return 1; if (n==1)return *this; int m=n; while (m>1) { t=*this; for ( i=1;i<<1<=m;i<<=1) { t=t*t; } m-=i; ret=ret*t; if (m==1)ret=ret*(*this); } return ret;}bool BigNum::operator>(const BigNum & T) const{ int ln; if (len > T.len) return true; else if (len == T.len) { ln = len - 1; while (a[ln] == T.a[ln] && ln >= 0) ln--; if (ln >= 0 && a[ln] > T.a[ln]) return true; else return false; } else return false;}bool BigNum::operator==(const BigNum & T) const{ int ln; if (len != T.len) return false; else { ln = len - 1; while (a[ln] == T.a[ln] && ln-- ); if (ln < 0) return true; else return false; }}bool BigNum::operator >(const int & t) const{ BigNum b(t); return *this>b;}bool BigNum::operator==(const int & t) const{ BigNum b(t); return *this == b;}/* 上面部分是模板辣 */const int _DP = 601;BigNum dp[_DP][_DP];BigNum tmp(1);int main() {for (int i = 0; i < _DP; ++i) dp[0][i] = tmp; //初始化dp[0][i]为1,为什么呢?其实也可以不用,但是下面就要j<i了for (int i = 1; i < _DP; ++i) {for (int j = 0; j <= (i>>1); ++j) {for (int k = 1; k <= j; ++k)dp[i][j] = dp[i][j] + dp[i-j][k];dp[i][0] = dp[i][0] + dp[i][j];}for (int j = (i>>1)+1; j <= i; ++j) {//也就是这里,就要j<i了,并且还要令dp[i][i] = 1,然后这个1也要记得加到dp[i][0]中去dp[i][j] = dp[i-j][0];dp[i][0] = dp[i][0] + dp[i][j];}}int n;ios::sync_with_stdio(false); //stream流加速,但是在toj上貌似效果不明显...while(cin >> n && n) {cout << dp[n][0] << endl;}return 0;}
当然,本题可以用大进制数,比如千万进制数,可以很方便地处理,并且可以节省很多时间,用类模板,毕竟省事一些...但是很慢,跑了1.01s,如果直接来个亿进制,可以控制到0.02s以内.这里不赘述了.
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