HPUOJ---2017寒假作业--专题-1/I-the Sum of Cube

I - the Sum of Cube

 

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

InputThe 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,BA,B. OutputFor 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 �C 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

思路：这道题关键是数据类型的转换，虽然题目范围A,B都是int型，但是立方后可能会变成long long,各项之

和相加得到的结果更有可能为long long,所以立方应设为long long,最后总和也应是long long;然而直接令总

和是long long，并不对，全部是int型的数据四则运算结果还是int,不会自动转化为long long,所以应令A或B

至少一个为long long,那样最终结果才会是long long;嗯，就这样。

代码一：

#include
int main()
{
	int T,k=0;
	long long A,B,i,sum=0;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lld%lld",&A,&B);
		for(i=A;i<=B;i++)
			sum+=i*i*i;
		k++;
		printf("Case #%d: %-lld\n",k,sum);
		sum=0;
	}
	return 0; 
}

代码二：运用数组，

#include
#include
#define max 1000000+11
long long a[max];
int main()
{
	memset(a,0,sizeof(a));
	int j;
	long long mul,i,SUM=0;
	for(i=1;i<=max;i++)
	{
		mul=1;
		for(j=1;j<=3;j++)
			mul*=i;
		SUM+=mul;
		a[i]=SUM;          //a[i]表示1到i所有数据三次方之和
	}
 	int T,k=0;
 	long long A,B;
 	scanf("%d",&T);
 	while(T--)
 	{
 		scanf("%lld%lld",&A,&B);
 		k++;
 		printf("Case #%d: %-lld\n",k,a[B]-a[A-1]);
	}
	 return 0;
 }