找單詞
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5842 Accepted Submission(s): 4099
然後包括N行數據,每行包括26個<=20的整數x1,x2,.....x26.
題目大意:單詞A-Z具有1-26的價值,現有字母A-Z的個數num[i],求問在不超過價值爲五十的情況下,有多少種字母的組合數;
解題思路:母函數;用指數代表價值,價值又爲數組的下標;用係數代表組成該價值的方案數,方案數爲數組中存的值;
代碼如下:
#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;
const int maxn = 50;
int num[maxn + 1];
int c1[maxn + 1], c2[maxn + 1];
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個多項式
if(num[i] == 0) continue;
for(int j = 0;j <= maxn;j ++){ //共有maxn+1項
for(int k = 0;k <= num[i] && j + k*i <= maxn;k ++)
c2[j + k*i] += c1[j];
}
for(int j = 0;j <= maxn;j ++){
c1[j] = c2[j];
c2[j] = 0;
}
}
int sum = 0;
for(int j = 1;j <= maxn;j ++)
sum += c1[j];
printf("%d\n",sum);
}
return 0;
}