首先想到的用回溯来求解,但是爆了超时,后来到网上看了解题报告,用到了一个新东西:母函数。
copy了大佬的代码,写在博客里,方便日后参考
自己写的回溯代码:
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #include<set> #include<cstring> #include<string> #include<iostream> #include<cmath> using namespace std; int a[30]; int cou=0; void f(int sum,int i){ if(i==26) return ; if(sum>50) return; if(a[i]==0){ f(sum,i+1); // printf("%d\n",sum); return ; } for(int j=0;j<=a[i];j++){ sum+=(i+1)*j; if(j!=0&&sum<=50&&sum!=0){ cou++; //printf("%d\n",sum); } f(sum,i+1); sum-=(i+1)*j; } } int main (){ int t; scanf("%d",&t); while(t--){ cou=0; for(int i=0;i<26;i++) scanf("%d",&a[i]); f(0,0); printf("%d\n",cou); } }网上大佬的母函数代码:
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #include<set> #include<cstring> #include<string> #include<iostream> #include<cmath> using namespace std; int num[55]; int c1[55], c2[55]; int main(){ int T; scanf("%d",&T); while(T --){ for(int i = 1;i <= 26;i ++) scanf("%d", &num[i]); memset(c1, 0, sizeof c1); memset(c2, 0, sizeof c2); for(int i = 0;i <= num[1];i ++) c1[i] = 1; for(int i = 2;i <= 26;i ++){ //共有26个多项式 if(num[i] == 0) continue; for(int j = 0;j <= 50;j ++){ //共有maxn+1项 for(int k = 0;k <= num[i] && j + k*i <= 50;k ++) c2[j + k*i] += c1[j]; } for(int j = 0;j <= 50;j ++){ c1[j] = c2[j]; c2[j] = 0; } } int sum = 0; for(int j = 1;j <= 50;j ++) sum += c1[j]; printf("%d\n",sum); } return 0; }