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主题
Calculate a + b
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杭电OJ-1000
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Input
Each line will contain two integers A and B. Process to end of file.
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Output
For each case, output A + B in one line.
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Mine
#include <stdio.h> int main() { int a,b; while(~scanf("%d %d",&a,&b)) //多次输入a和b。 { printf("%d ",a+b); } } /*** while(~scanf("%d %d",&a,&b)) 多次输入a和b。 这句话中的“~”符号可以理解为“重复”，代码含义是反复执行scanf(“%d %d”,&a,&b) 语句，直到语句接收不到有效结果。换一种说法就是while语句会在括号中的判断为真的情况执行语句，那么对于scanf函数而言，判断为真也就是接收到了有效数据。而~符号代表无限重复，直到scanf语句不能取到有效的值为止（while的括号中判断为假），循环跳出。 ***/ -
Review
题目: 接收两个整数并返回两个数的和。
需要注意的是题目中说明了每行两个数据，但并没有说明多少行。换一种常用说法叫：“多组数据”，是常见的要求。但没有C语言算法书会写明接收多组数据的方式。
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杭电OJ-1001
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Description
calculate SUM(n) = 1 + 2 + 3 + ... + n.
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Input
The input will consist of a series of integers n, one integer per line.
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Outpu
For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.
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Mine
#include<stdio.h> int main() { int x,n,sum; scanf("%d",&x); scanf("%d",&n); for(x=1;x<=n;x++) { sum=0; sum=sum+x; } printf("1 %d",sum); } -
Writeup
#include <stdio.h> int main() { int a; int sum=0; while((scanf("%d",&a))!=EOF){ for(int i=0;i<=a;i++) sum = sum+i; printf("%d ",sum); sum = 0; } return 0; } -
Review
- 我的问题在于没有考虑连续读取的可能性，好像对输入输出有误解T-T
- 有一个疑问：输入输出需要和sample一样的格式吗？
- 本身是一个前N项和的累加问题，如果用公式法也是no accept,原因是S=(1+n)*n/2中乘法容易造成溢出，而循环累加的好处在于溢出的可能比较小。
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杭电OJ-1002
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Description
Given two integers A and B, your job is to calculate the Sum of A + B.
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Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
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Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
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手写代码@V1-20210805
#include<stdio.h> #include<string.h> #define max 1005 int main() { int a[max],b[max],c[max]; while(~scanf("%s1 %s2",s1,s2)) { int i,j,k; i=0；j=0; k=strlen(s1)>strlen(s2)?strlen(s1):strlen(s2) for(i;i<=k;i++) //k有无定义的必要？ { a[]=s1; //好像不太合适？字符数组能直接赋值给数组吗？查一下书 b[]=s2; //似乎需要循环读入？ c[i]=a[i]+c[i]; if(c[i]>=10) { c[i]=c[i]%10; c[i+1]++; //s2的长度是必要的吗？ } } for(j=0;j<=k;j++) { printf("%d",a[j]); //哪里有点奇怪 } } } -
v1现在存在的问题
- 读入的顺序和相加的顺序不太对？要处理一下？[n-i]好像可行？
- 输出的时候应该是逆序？
- 需要加一个读入的限制条件：读取正整数？
- 题目要求的输入输出 用一个循环？
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手写代码@V2-20210806
#include<stdio.h> #include<string.h> #define max 1010 int main() { int T,u,n,i,j; char s1[max],s2[max],c[max]; scanf("%d",&T); while(T>=1 && T<=20) { for(u=0;u<=T;u++) { scanf("%s %s",s1,s2); } n =strlen(s1)>strlen(s2)?strlen(s1):strlen(s2); for(i=2 ; i<=n ; i++) { c[n-i]=s1[n-i]+s2[n-i]; if(c[i]>=10) { c[n-i]=c[n-i]%10; c[n-i-1]++; //没有考虑不是同位数的情况 } } for(j=0;j<=n;j++) { printf("case %d: ",u); printf("%s + %s = %d ",s1,s2,c[j]); } } } -
writeup-0806
#include<stdio.h> #include<string.h> #define max 1000+10 /** * 1. define the variable **/ int a[max],b[max]; char str1[max],str2[max]; int main(){ int m; //test number T? int k=1; scanf("%d",&m); // read T? /** *2.read number and make sure input. this part is to read the big number and covert to array **/ while(m--){ //this circle ie funny! it's better than mine which use more variable u. scanf("%s %s",str1,str2); //read big number memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); //memset() 函数可以说是初始化内存的“万能函数”，通常为新申请的内存进行初始化工作。 int i,j; for(i=0,j=strlen(str1)-1;i<strlen(str1);i++){ //it's similar to my code ,i use the i=1 to replace len a[j--]=str1[i]-'0'; //string1 covert to array1? } for(i=0,j=strlen(str2)-1;i<strlen(str2);i++){ //two strlen conditions,less code ,nice~ b[j--]=str2[i]-'0'; //string2 to array2? why not combine with above 'for cicle' together ? } /** * 3. add two big number **/ for(i=0;i<max;i++){ a[i]+=b[i]; if(a[i]>=10){ a[i]-=10; //mine : a[i]=a[i]%10 a[i+1]+=1; //similar~ } } /** * 4.output * **/ printf("Case %d: ",k++); // k has been defined and value=1? printf("%s + %s = ",str1,str2); for(i=max-1;(i>=0)&&(a[i]==0);i--); //reverse output？(i>=0)&&(a[i]==0) what's mean? only deal the 10? if(i>=0){ for(;i>=0;i--){ printf("%d",a[i]); } } else printf("0"); if(m!=0) printf(" "); else printf(" "); //not clear.. } return 0; } -
Review
- 大数加法问题一般考虑数组进行存储，然后按位相加满十进一。
- 再看Writeup时发现大家再提java,
import java.util.Scanner，以及大数需要import java.math.BigInteger,且BigInterger相加不是"a+b"，而是"a.add(b)",就可以很好的解决大数问题。
import java.util.Scanner; import java.math.BigInteger; public class Main{ public static void main(String args[]){ BigInteger a,b; int T; int n=1; Scanner in = new Scanner(System.in); T=in.nextInt(); while(T>0){ a=in.nextBigInteger(); b=in.nextBigInteger(); System.out.println("Case "+n+":"); System.out.println(a+" + "+b+" = "+a.add(b)); if(T!=1) System.out.println(); T--; n++; } } } -
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杭电OJ-1089
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Input
The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.
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Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
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**Mine **- accepted
#include<stdio.h> int main() { int a,b; while(~scanf("%d %d",&a,&b)) { printf("%d ",a+b); } return 0; }
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杭电OJ-1090
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Input
Input contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.
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Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
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Mine
#include<stdio.h> int i,a,b,c; int main() { while(~scanf("%d",&i)) { for(c=1;c<=i;c++) { scanf("%d %d",&a,&b); printf("%d ",a+b); } } }
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杭电OJ-1091
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Input
Input contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.
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Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
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Mine
#include<stdio.h> int a,b; int main() { while(~scanf("%d %d",&a,&b)) { if(a==0 && b==0) {return 0;} else { printf("%d ",a+b); } } }
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杭电OJ-1092
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Input
Input contains multiple test cases. Each test case contains a integer N, and then N integers follow in the same line. A test case starting with 0 terminates the input and this test case is not to be processed.
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Output
For each group of input integers you should output their sum in one line, and with one line of output for each line in input.
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Mine
#include<stdio.h> //始终是wrong... int main() { int a,n,sum; while(scanf("%d",&n)!=EOF && n!=0) { while(n--) { scanf("%d",&a); sum=0; sum+=a; } printf("%d ",sum); } return 0; } -
Writeup
#include<stdio.h> int main() { int a[10000]; int n, i, s; while (scanf("%d", &n) && n) //输入正确scanf返回1，n!=0继续输入 { for (i = s = 0; i < n; i++) scanf("%d", &a[i]); for (i = 0; i < n; i++) { s = s + a[i]; } printf("%d ", s); } return 0; } -
Review
没有考虑到大数的可能，存在溢出问题，用数组存放更为合理
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杭电OJ-1093
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Input
Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.
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Output
For each group of input integers you should output their sum in one line, and with one line of output for each line in input.
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Mine
#include<stdio.h> //wrong answer... int main() { int i,j,k,s; int a[1000]; while(~scanf("%d",&i)) { for(i;i>=1;i--) { while(scanf("%d",&j)!=EOF && j) { for(j;j>=0;j--) { scanf("%d", &a[i]); } for (k= 0; k< j; k++) { s += a[i]; } printf("%d ", s); } } } } -
writeup
#include<stdio.h> int main() { int n,i,m,sum; //n是行数，m是加数个数 scanf("%d",&n); while(n--) //简洁！ { sum=0; scanf("%d",&m); while(m--) //两个-- 简洁欸！ { scanf("%d",&i); sum=sum+i; } printf("%d ",sum); } return 0; } -
Review
所以上一道题不是大数的原因？？ (((φ(◎ロ◎;)φ)))
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杭电OJ-1094
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Input
Input contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line.
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Output
For each test case you should output the sum of N integers in one line, and with one line of output for each line in input.
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Mine
#include<stdio.h> int main() { int n,m,sum; while(~scanf("%d",&n)) { sum = 0; while(n--) { scanf("%d",&m); sum += m; } printf("%d ",sum); } }
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杭电OJ-1095
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Input
The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.
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Output
For each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.
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Mine
#include<stdio.h> int main() { int a,b; while(~scanf("%d %d",&a,&b)) { printf("%d ",a+b); printf(" "); } return 0; }
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杭电OJ-1096
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Input
Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.
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Output
For each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.
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Mine
#include<stdio.h> int main() { int n,m,i; int sum; while(~scanf("%d ",&n)) //行数 { while(n--) { scanf("%d",&m); //加数个数 sum = 0; while(m--) { scanf("%d",&i); sum += i; } printf("%d ",sum); if(m!=0) printf(" ",sum); } } }
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