2017年ICPC中国大陆区域赛真题（下）E题

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本文链接： https://blog.csdn.net/qq_43735840/article/details/99730871

题目网址：点击进入

problem:
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

大致题意：
输入t组数据每组数据四个数，求和

思路：眼看是水题，再看看数据范围如果极限的话结果和爆范围了，所以需要处理一下数字，或许用模拟高精度加法也可以

代码：

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
typedef unsigned long long ll;
int main()
{
    int t;
    cin>>t;
    while (t--)
    {
        ll a,b,c,d;
        cin>>a>>b>>c>>d;///极限数据下-1使得范围在ll里面，把个位拿出来单独输出
        if (a==(1ll<<62)&&b==(1ll<<62)&&c==(1ll<<62)&&d==(1ll<<62))
            cout<<(a-1+b+c+d)/10<<(a-1+b+c+d)%10+1<<endl;
        else
            cout<<a+b+c+d<<endl;
    }
    return 0;
}

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
typedef long long LLD;
LLD sum[1005];///总和
char a[1005],b[1005],c[1005],d[1005];///四个数字字符串输入
int main()
{
    LLD t,length,MAX=0,first;///t组，数字长度，四个数字最长，反序指针标记
    scanf("%lld%*c",&t);
    while (t--)
    {
        memset(a,0,sizeof(a));///所有全部归0
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));
        memset(sum,0,sizeof(sum));
        scanf("%s%*c%s%*c%s%*c%s%*c",&a,&b,&c,&d);
        length=strlen(a),first=0;
        for (LLD i=length-1;i>=0;i--)
        {
            sum[first]=sum[first]+(a[i]-'0');///从末尾个位数开始相加
            if (sum[first]>=10)
            {
                sum[first]=sum[first]%10;///大于10就进位处理
                sum[first+1]++;
            }
            first++;
        }
        MAX=max(MAX,first);///计算最大长度
        length=strlen(b),first=0;
        for (LLD i=length-1;i>=0;i--)
        {
            sum[first]=sum[first]+(b[i]-'0');
            if (sum[first]>=10)
            {
                sum[first]=sum[first]%10;
                sum[first+1]++;
            }
            first++;
        }
        MAX=max(MAX,first);
        length=strlen(c),first=0;
        for (LLD i=length-1;i>=0;i--)
        {
            sum[first]=sum[first]+(c[i]-'0');
            if (sum[first]>=10)
            {
                sum[first]=sum[first]%10;
                sum[first+1]++;
            }
            first++;
        }
        MAX=max(MAX,first);
        length=strlen(d),first=0;
        for (LLD i=length-1;i>=0;i--)
        {
            sum[first]=sum[first]+(d[i]-'0');
            if (sum[first]>=10)
            {
                sum[first]=sum[first]%10;
                sum[first+1]++;
            }
            first++;
        }
        MAX=max(MAX,first);///如果最高位数0的话就-1防止出现0122开头0的结果
        if (sum[MAX]==0)
        {
            MAX--;
        }
        for (LLD i=MAX;i>=0;i--)///从最高位输出
        {
            printf("%d",sum[i]);
        }
        printf("\n");
    }
    return 0;
}