Little Boxes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2374 Accepted Submission(s): 845
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.
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.
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
/* 题意: 给n组数据,每组四个数,将四个数之和输出即可 思路: long long 会溢出,简单大数加法模拟。 */ #include<bits/stdc++.h> using namespace std; int bit[100],next1[100]; /* @function: 模拟计算 @param: next1[]:记录每位值的大小 bit[] :记录加法后的结果 */ void cul(char *x){ int len=strlen(x); reverse(x,x+len); //倒置数组 memset(next1,0,sizeof next1); for(int i=0;i<len;i++) next1[i]=x[i]-'0'; for(int i=0;i<len;i++){ bit[i]+=next1[i]; while(bit[i]>=10){ bit[i+1]++; bit[i]-=10; } } } int main(){ char a[100],b[100],c[100],d[100]; int n; scanf("%d",&n); for(int i=0;i<n;i++){ memset(bit,0,sizeof bit); cin>>a>>b>>c>>d; cul(a); cul(b); cul(c); cul(d); int f=0; //处理前置0 for(int i=99;i>=0;i--){ if(bit[i]>0){ f=i; break; } } for(int i=f;i>0;i--){ printf("%d",bit[i]); } printf("%d\n",bit[0]); } }