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C語言中如何實現多組數據輸入輸出？
您好：#include iostream
#include stdlib.h
using namespace std;
int main()
{
int n;
int a[50000];
while (cinn) //當沒有n輸入的時候結束循環，可以按 ctrl+z 來輸入結束符EOF
{
for (int i=0;i50000;i++)
a[i]=0;
for (int i=0;in;i++)
{
int temp;
cintemp;
a[temp]=temp;
}
for (int i=0;i50000;i++)
{
if (a[i] != 0)
couta[i]” “;
}
}
system(“pause”);
return 0;
}
追問
先謝謝你。但是這段代碼不是我想要的，我也寫過。
當輸完第一組：
5
1 2 5 4 5
回車之後，馬上輸出第一組的結果：
1 2 4 5
我想線不輸出第一組的結果，等我把第二個case輸進去之後，按ctrl + Z 後再輸出兩個case的結果。
C語言如何多組數據輸入輸出
#includeintpow(inta,intn)//計算a的n次方{if(n==1)returna;returna*pow(a,n-1);}intmain(){intT;intn,k,sum,i;scanf(“%d”,T);while(T–){sum=0;scanf(“%d%d”,n,k);for(i=1;i
C語言中如何實現多組數據輸入輸出
仔細認真看看下面的會對你有幫助的，嘿嘿
輸入格式：有多個case輸入，直到文件結束
輸出格式：一行一個結果
Problem Description
Your task is to Calculate a + b.
Too easy?! Of course! I specially designed the problem for acm beginners.
You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.
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.
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.
Sample Input
1 5
10 20
Sample Output
6
30
Author
lcy
Recommend
JGShining
#include stdio.h
int main()
{
int a,b;
while( scanf( “%d%d” , a , b ) != EOF ) //輸入直到文件結尾
{
printf( “%d\n” , a+b ); //一行一個結果
}
return 0;
}
HDOJ1090
輸入格式：先輸入有case數，再依次輸入每個case
輸出格式：一行一個結果
#include stdio.h
Problem Description
Your task is to Calculate a + b.
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.
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.
Sample Input
2
1 5
10 20
Sample Output
6
30
Author
lcy
Recommend
JGShining
int main()
{ int n,a,b;
scanf( “%d” , n ); //輸入的case數
while( n– ) //控制輸入
{ scanf( “%d%d” , a , b );
printf( “%d\n” , a+b ); //一行一個結果
}
return 0;
}
HDOJ1091
輸入格式：每行輸入一組case，當case中的數據滿足某種情況時退出
輸出格式：一行一個結果
Problem Description
Your task is to Calculate a + b.
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.
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.
Sample Input
1 5
10 20
0 0
Sample Output
6
30
Author
lcy
Recommend
JGShining
#include stdio.h
int main()
{
int a,b;
while( scanf( “%d%d” , a , b ) (a||b) ) //輸入直到滿足a和b均為0結束
{
printf( “%d\n” , a+b ); //一行一個結果
}
return 0;
}
HDOJ1092
輸入格式：每組case前有一個控制輸入個數的數，當這個數為0結束
輸出格式：一行一個結果
#include stdio.h
Problem Description
Your task is to Calculate the sum of some integers.
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.
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.
Sample Input
4 1 2 3 4
5 1 2 3 4 5
Sample Output
10
15
Author
lcy
Recommend
JGShining
int main()
{
int n,sum;
while( scanf( “%d” , n ) n ) //每組case前有一個控制該組輸入數據的數，為0結束
{
int x;
sum = 0;
while( n– ) //控制該組輸入個數
{
scanf( “%d” , x );
sum += x;
}
printf( “%d\n” , sum ); //一行一個結果
}
return 0;
}
HDOJ1093
輸入格式：一開始有一個控制總的輸入case的數，而每個case中又有一個控制該組輸入數據的數
輸出格式：一行一個結果
Problem Description
Your task is to calculate the sum of some integers.
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.
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.
Sample Input
2
4 1 2 3 4
5 1 2 3 4 5
Sample Output
10
15
Author
lcy
5
#include stdio.h
int main()
{
int casnum,n,sum;
scanf( “%d” , casnum ); //控制總的輸入case的數
while( casnum– ) //控制總的輸入個數
{
int x;
sum = 0;
scanf( “%d” , n ); //每個case中控制該組輸入個數
while( n– )
{
scanf( “%d” , x );
sum += x;
}
printf( “%d\n” , sum ); //一行一個結果
}
return 0;
}
HDOJ1094
輸入格式：總的case是輸到文件結尾，每個case中的一開始要輸入一個控制該組個數的數
輸出格式：一行一個結果
Problem Description
Your task is to calculate the sum of some integers.
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.
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.
Sample Input
4 1 2 3 4
5 1 2 3 4 5
Sample Output
10
15
6
#include stdio.h
int main()
{
int n,sum;
while( scanf( “%d” , n ) != EOF ) //輸出到文件結尾
{
int x;
sum = 0;
while( n– ) //控制該組輸入個數
{
scanf( “%d” , x );
sum += x;
}
printf( “%d\n” , sum ); //一行一個結果
}
return 0;
}
HDOJ1095
輸入格式：輸入直到文件結束
輸出格式：一行一個結果，結果輸完後還有一個blank line
Problem Description
Your task is to Calculate a + b.
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.
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.
Sample Input
1 5
10 20
Sample Output
6
30
7
#include stdio.h
int main()
{
int a,b;
while( scanf( “%d%d” , a , b ) != EOF ) //輸入直到文件結束
{
printf( “%d\n\n” , a+b ); //一行一個結果，結果輸完後還有一個回車
}
return 0;
}
HDOJ1096
輸入格式：一開始輸入總的case數，每組case一開始有控制該組輸入個數的數
輸出格式：一行一個結果，兩個結果之間有一個回車，注意最後一個case的處理。
Problem Description
Your task is to calculate the sum of some integers.
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.
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.
Sample Input
3
4 1 2 3 4
5 1 2 3 4 5
3 1 2 3
Sample Output
10
15
6
#include stdio.h
int main()
{
int casnum,n,sum;
scanf( “%d” , casnum ); //總的輸入case數
while( casnum– ) //控制輸入組數
{
int x;
sum = 0;
scanf( “%d” , n ); //控制每組的輸入個數
while( n– )
{
scanf( “%d” , x );
sum += x;
}
printf( “%d\n” , sum ); //一行一個結果
if( casnum ) printf( “\n” ); //兩兩結果之間有一個回車，最後一個結果後面沒有
}
return 0;
}
c語言中，一次連續輸入多組數據，並且最後連續輸出多組結果，應該用哪種方法
用二維數組就可以實現一次連續輸入多組數據。思路是嵌套循環，外層循環控制二維數組的行數（也就是第幾組數據），內層循環控制這組數據中數據個數。
採用二維數組方法的有點在於，這種隨機存取的數據結構方便查找和檢索，但一定要注意這種方法不便於向已有數據中插入和刪除數據。
原創文章，作者：小藍，如若轉載，請註明出處：https://www.506064.com/zh-hk/n/240166.html
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