HDU1028 Ignatius and the Princess III【母函式】【完全揹包】
題目連結:
題目大意:
給定正整數N,定義N = a[1] + a[2] + a[3] + … + a[m],a[i] > 0,1 <= m <= N。
對於給定的正整數N,問:能夠找出多少種這樣的等式?
思路:
對於N = 4,
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1。
共有5種。N=4時,結果就是5。其實就是整數分解問題,可寫出母函式
g(x) = (1+x+x^2+x^3+…)*(1+x^2+x^4+…)*(1+x^3+…)*(1+x^4+…)
出解即可。
其實,用完全揹包也可以做,附上程式碼。
AC程式碼:
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> using namespace std; int c1[130],c2[130]; int main() { int N; while(cin >> N) { for(int i = 0; i <= N; ++i) { c1[i] = 1; c2[i] = 0; } for(int i = 2; i <= N; ++i) { for(int j = 0; j <= N; ++j) for(int k = 0; j+k <= N; k += i) c2[j+k] += c1[j]; for(int j = 0; j <= N; ++j) { c1[j] = c2[j]; c2[j] = 0; } } cout << c1[N] << endl; } return 0; }
#include<stdio.h> #include<string.h> int dp[200]; int main() { dp[0] = 1; for(int i = 1; i <= 120; i++) { for(int j = i; j <= 120; j++) { dp[j] += dp[j-i]; } } int n; while(~scanf("%d",&n)) { printf("%d\n",dp[n]); } return 0; }
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