poj1028 Ignatius and the Princess III hdu1709The Balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9592    Accepted Submission(s): 6768 Problem 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[i]>0,1<=m<=N; My question is how many different equations you can find for a given N. For example, assume N is 4, we can find:   4 = 4;   4 = 3 + 1;   4 = 2 + 2;   4 = 2 + 1 + 1;   4 = 1 + 1 + 1 + 1; so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"   Input The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.   Output For each test case, you have to output a line contains an integer P which indicate the different equations you have found.   Sample Input 4 10 20   Sample Output 5 42 627   Author Ignatius.L 经典的母函数题，详见代码用好这一们模板，其实很多母函数就是在套用而矣！ #include <iostream> #include <stdio.h> using namespace std; int tempresult[121],result[121]; int main() { int n,i,j,k; while(scanf("%d",&n)!=EOF) { for(i=0;i<=n;i++)//初始化 { result[i]=1;//一个式子一个式子的乘积所得到的系数,最开始为(1+x+x^2+x^3...)所以全为１ tempresult[i]=0;//这个就是在计算中临时保存结果的数列，最终还原给result } for(i=2;i<=n;i++)//i代表第i个括号里的式子如（１+x^2+x^4...)所对应的i为２以此向后推 { for(j=0;j<=n;j++)//j代表是，每一次算出来的结果式子的系数,这里，因为只求n最大的指数，所以可以把大于n的不看了，因为， 已经大于n了，那么算的结果，对最后的结果，不再产生影响 for(k=0;k<=n-j;k=k+i)//k表示的是当前要相乘的数列的指数，如第一次要和，(1+x^2+x^3+...),第二次，要和(1+x^3+x^6...), 我们可以看出，k这三个数列，每次指数，加的都是1,2,3,这不就是加的i么,和上面一样，大于n,再算对最后的结果没影响，自然可以不计 { tempresult[j+k]+=result[j];//新形成的数列的指数为j+k的系数，因为，新乘的项系数都为1,故只用加个result[j], 就可以了 } for(j=0;j<=n;j++) { result[j]=tempresult[j];//把乘了一项了系数都保存起来 tempresult[j]=0;//临时保存的没有用了，清0 } } printf("%d\n",result[n]); } return 0; } 我们再看一个，非常相似的问题 Home Problems Status Contest     62:12:11 384:00:00 Overview Problem Status Rank A  B  C  D  E  F  G  H  I  J  K  L  M  N  O A - The Balance Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit  Status Description Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.   Input The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.   Output For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.   Sample Input 3 1 2 4 3 9 2 1   Sample Output 0 2 4 5 在这个问题中，我们有几个重量，我们要求的是，什么不能称出， 其实，这也是可以用母函数做的！ #include<stdio.h> #include<math.h> int c2[100000],c1[100000]; int main() { int i,j,k,s,m,x,n,a[105]; while(scanf("%d",&n)!=EOF) { for(i=0,s=0;i<n;i++) { scanf("%d",&a[i]); s+=a[i]; } for(i=0;i<s;i++) { c2[i]=0; } for(i=0,k=0,m=0;i<n;++i) { if(c2[a[i]-1]==0) { c1[m]=a[i]; c2[a[i]-1]++; m++; }//printf("%d ",j); for(j=0;j<k;++j) { x=a[i]+c1[j]; if(c2[x-1]==0) { c1[m]=x; c2[x-1]++; m++; } x=abs(a[i]-c1[j]); if(c2[x-1]==0&&x>0) { c1[m]=x; c2[x-1]++; m++; } } k=m; } //for(i=0;i<k;i++) // printf("%d ",c1[i]); printf("%d\n",s-k); if(k!=s) { for(j=0,i=0;i<s;i++) { if(c2[i]==0) if(j!=0) printf(" %d",i+1); else { j++; printf("%d",i+1); } } printf("\n"); } } return 0; }   FAQ | About Virtual Judge |  Forum |  Discuss |  Open Source Project All Copyright Reserved ©2010-2012  HUST ACM/ICPC TEAM  Anything about the OJ, please ask in the  forum, or contact author: Isun Server Time:    Statistic |  Submit |  Discuss |  Note

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