ZOJ Problem Set - 2773 Triangular Sums【公式】
2016-03-01 13:42
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ZOJ Problem Set - 2773
Triangular Sums
Time Limit: 2 Seconds
Memory Limit: 65536 KB
The nth Triangular number, T(n) = 1 + ... + n, is the sum of the first n integers. It is the number of points in a triangular array with n points on side.
For example T(4):
Write a program to compute the weighted sum of triangular numbers:
W(n) = SUM[k = 1..n; k*T(k+1)]
Input
The first line of input contains a single integer N, (1 <= N <= 1000) which is the number of datasets that follow.
Each dataset consists of a single line of input containing a single integer n, (1 <= n <= 300), which is the number of points on a side of the triangle.
Output
For each dataset, output on a single line the dataset number, (1 through N), a blank, the value of n for the dataset, a blank, and the weighted sum , W(n), of triangular numbers for n.
Sample Input
4
3
4
5
10
Sample Output
1 3 45
2 4 105
3 5 210
4 10 2145
Source: Greater New York Regional 2006
解题思路:
两个公式的递推。
Triangular Sums
Time Limit: 2 Seconds
Memory Limit: 65536 KB
The nth Triangular number, T(n) = 1 + ... + n, is the sum of the first n integers. It is the number of points in a triangular array with n points on side.
For example T(4):
X X X X X X X X X X
Write a program to compute the weighted sum of triangular numbers:
W(n) = SUM[k = 1..n; k*T(k+1)]
Input
The first line of input contains a single integer N, (1 <= N <= 1000) which is the number of datasets that follow.
Each dataset consists of a single line of input containing a single integer n, (1 <= n <= 300), which is the number of points on a side of the triangle.
Output
For each dataset, output on a single line the dataset number, (1 through N), a blank, the value of n for the dataset, a blank, and the weighted sum , W(n), of triangular numbers for n.
Sample Input
4
3
4
5
10
Sample Output
1 3 45
2 4 105
3 5 210
4 10 2145
Source: Greater New York Regional 2006
解题思路:
两个公式的递推。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; double oo(int i) { return (i*i*i+3*i*i+2*i)/2; } int main() { int T; int y=1; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); printf("%d %d ",y++,n); int i,j; double ans=0; for(i=1;i<=n;i++) { ans+=oo(i); } printf("%.0lf\n",ans); } return 0; }
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