HDU —— 6225

Little Boxes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1779    Accepted Submission(s): 664

Problem Description

Little boxes on the hillside.
Little boxes made of ticky-tacky.
Little boxes.
Little boxes.
Little boxes all the same.
There are a green boxes, and b pink boxes.
And c blue boxes and d yellow boxes.
And they are all made out of ticky-tacky.
And they all look just the same.

 

Input

The input has several test cases. The first line contains the integer t (1 ≤ t ≤ 10) which is the total number of test cases.
For each test case, a line contains four non-negative integers a, b, c and d where a, b, c, d ≤ 2^62, indicating the numbers of green boxes, pink boxes, blue boxes and yellow boxes.

 

Output

For each test case, output a line with the total number of boxes.

 

Sample Input

4 1 2 3 4 0 0 0 0 1 0 0 0 111 222 333 404

 

Sample Output

10 0 1 1070

 

这题把我搞死。。。满心欢喜的觉得超简单很快能过，

第一遍long long 直接加 wa,,,,,

第二遍unsighned long long     wa

第三遍改用大数，，，wa

第四遍大数修改，，，wa（再改一点就过）

第五遍想了一个超级超级巧妙的方法，贴错了代码。。。错了代码。。。代码。。。（这种方法可以过的）

第六遍，，队友过（论有个好队友的重要性）

可长点心吧，这题耽误时间我的锅。。。

方法一：

思路：unsigned long long ,所以至少可以用两个long long 分别加两个数，话不多说贴代码

#include<bits/stdc++.h>
using namespace std;
const int MAX = 80;
int x[MAX];
int y[MAX];
int ans[MAX];
int main()
{
    int t;
    unsigned long long a, b, c, d, temp1, temp2;
    while(scanf("%d", &t) != EOF)
    {
        while(t--)
        {
            memset(ans, 0, sizeof(ans));
            memset(x, 0, sizeof(x));
            memset(y, 0, sizeof(y));
            scanf("%lld %lld %lld %lld", &a, &b, &c, &d);
            temp1 = a+b;
            temp2 = c+d;
            int cnt1 = 79, cnt2 = 79;
            while(temp1)
            {
                x[cnt1--] = temp1%10;
                temp1 /= 10;
            }
            while(temp2)
            {
                y[cnt2--] = temp2%10;
                temp2 /= 10;
            }
            int temp = 79;
            while(temp >= cnt1 && temp >= cnt2)
            {
                if(x[temp] + y[temp] >= 10 - ans[temp])
                {
                    ans[temp] = (ans[temp] + x[temp] + y[temp])%10;
                    ans[temp-1]++;
                }
                else
                    ans[temp] += (x[temp] + y[temp]);
                temp--;
            }
            while(temp >= cnt1)
            {
                if(x[temp] + ans[temp] > 9)
                {
                    ans[temp] = (x[temp] + ans[temp])%10;
                    ans[temp-1]++;
                }
                else
                    ans[temp] += x[temp];
                temp--;
            }
            while(temp >= cnt2)
            {
                if(y[temp] + ans[temp] > 9)
                {
                    ans[temp] = (y[temp] + ans[temp])%10;
                    ans[temp-1]++;
                }
                else
                    ans[temp] += y[temp];
                    temp--;
            }
            while(ans[temp] == 0 && temp != 79)
                temp++;
            for(int i = temp; i <= 79; i++)
            {
                cout<<ans[i];
            }
            cout<<endl;
        }
    }
    return 0;
}

方法二：

思路：四个数相加最大是2^64,unsighned long long 最大为2^64-1，所以当四个数全是2^62时

输出2^64，其他情况正常相加后输出

按例贴代码

#include<bits/stdc++.h>
using namespace std;
const int MAX = 80;
int x[MAX];
int y[MAX];
int ans[MAX];
int main()
{
	int t;
	unsigned long long a, b, c, d, temp1, temp2;
	scanf("%d", &t);
	while(t--)
	{
		scanf("%lld %lld %lld %lld", &a, &b, &c, &d);
		temp1 = a+b;
		temp2 = c+d; 
		if(a ==  4611686018427387904 && b == 4611686018427387904 && c == 4611686018427387904 && d == 4611686018427387904)
			cout<<"18446744073709551616"<<endl;
		else
			cout<<temp1 + temp2<<endl;
	}
	return 0;
}