PAT甲级 1001. A+B Format (20)--两种方法
2017-08-01 08:17
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题目链接:https://www.patest.cn/contests/pat-a-practise/1001
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input
Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
Output
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input
Sample Output
方法一:
通用方法。把两数和作为字符串处理。
方法二:
针对本题最多七位数的情况,写个特别的。
1001. A+B Format (20)
时间限制400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input
Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
Output
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input
-1000000 9
Sample Output
-999,991
方法一:
通用方法。把两数和作为字符串处理。
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; int main(){ int a,b; cin>>a>>b; int c=a+b; if(c<0) cout<<'-'; c=abs(c); char s[20]; sprintf(s,"%d",c); int len=strlen(s); //if(len<=3) {cout<<s;return 0;} int m=len/3,n=len%3,start=0; //cout<<"m="<<m<<' '<<"n="<<n<<endl; if(n==0) {cout<<s[0]<<s[1]<<s[2];start=3;m--;} else if(n==1) {cout<<s[0];start=1;} else if(n==2) {cout<<s[0]<<s[1];start=2;} while(m!=0){ cout<<','; cout<<s[start]<<s[start+1]<<s[start+2]; start+=3; m--; } return 0; }
方法二:
针对本题最多七位数的情况,写个特别的。
#include<iostream> #include<cstdio> using namespace std; int main(){ int a,b; cin>>a>>b; int c=a+b; if(c<0){cout<<'-';c=-c;} if(c>=1000000){ printf("%d,%03d,%03d",c/1000000,c%1000000/1000,c%1000); }else if(c>=1000){ printf("%d,%03d",c/1000,c%1000); }else{ printf("%d",c); } return 0; }
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