HDOJ 5053 the Sum of Cube (立方和)

the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2420    Accepted Submission(s): 1016

Problem Description

A range is given, the begin and the end are both integers. You should sum the cube of all the integers in the range.

 

Input

The first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve.
Each case of input is a pair of integer A,B(0 < A <= B <= 10000),representing the range[A,B].

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then output the answer – sum the cube of all the integers in the range.

 

Sample Input

  

   2
1 3
2 5
  

 

Sample Output

  

   Case #1: 36
Case #2: 224
  

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

题意：就是求解给定范围数的立方的和。

代码1：直接计算，注意数据类型

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define MYDD 1103

using namespace std;

int main() {
	int t,t2=1;
	double a,b,ans;
	scanf("%d",&t);
	while(t--) {
		scanf("%lf%lf",&a,&b);
		ans=0.0;
		for(int j=a; j<=b; j++)
			ans+=(double)pow(j,3);
		printf("Case #%d: %.0lf\n",t2,ans);
		t2++;
	}
	return 0;
}

代码，利用自然数立方和的公式：

S(n)=(n*(n+1))^2/4

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define MYDD 1103

using namespace std;

double AND(double x ) {
	return pow(x*(x+1),2)/4;
}

int main() {
	int t,t2=1;
	double a,b,ans;
	scanf("%d",&t);
	while(t--) {
		scanf("%lf%lf",&a,&b);
		ans=AND(b)-AND(a-1);
		printf("Case #%d: %.0lf\n",t2,ans);
		t2++;
	}
	return 0;
}