Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12521    Accepted Submission(s): 8838 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25805    Accepted Submission(s): 17839 Problem Description "Well, it seems the first problem is too easy. I will let…

这是道典型的母函数的题目,可以看看我的母函数这一标签上的另一道例题,里面对母函数做了较为详细的总结.这题仅贴上代码: #include"iostream" using namespace std; #define N 130 ],b[N+]; int main() { int n,i,j,k; ) { ;i<=n;i++) {a[i]=;b[i]=;} ;i<=n;i++) { ;j<=n;j++) ;k+j<=n;k+=i) { b[k+j]+=a[j]; }…

题意 给出$n$,问用$1$到$n$的数字问能构成$n$的方案数 思路 生成函数基础题,$x^{n}$的系数即答案. 代码 #include <bits/stdc++.h> #define DBG(x) cerr << #x << " = " << x << endl; using namespace std; const int N = 120 + 5; int n, c[2][N]; int main() { while(…

先咕着 ---------------2018 5 22---------------------- 题解 生成函数处理整数拆分 code #include<cstdio> #include<cstring> #include<algorithm> inline int raed() { int x = 0,f = 1; char c = getchar(); while(c < '0' || c < '9') c = getchar(); while(c…

大意是给你1个整数n,问你能拆成多少种正整数组合.比如4有5种: 4 = 4;  4 = 3 + 1;  4 = 2 + 2;  4 = 2 + 1 + 1;  4 = 1 + 1 + 1 + 1; 然后就是母函数模板题……小于n的正整数每种都有无限多个可以取用. (1+x+x^2+...)(1+x^2+x^4+...)...(1+x^n+...) 答案就是x^n的系数. #include<cstdio> #include<cstring> using namespace std;…

Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. "The second problem is, given an positive integer N, we define an equation like this:   N=a[1]+a[2]+a[3]+...+a[m];   a…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11810    Accepted Submission(s): 8362 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. &…

题目链接:hdu 1028 Ignatius and the Princess III 题意:对于给定的n,问有多少种组成方式 思路:dp[i][j],i表示要求的数,j表示组成i的最大值,最后答案是dp[i][i].那么dp[i][j]=dp[i][j-1]+dp[i-j][i-j],dp[i][j-1]是累加1到j-1的结果,dp[i-j][i-j]表示的就是最大为j,然后i-j有多少种表达方式啦.因为i-j可能大于j,这与我们定义的j为最大值矛盾,所以要去掉大于j的那些值 /*******…