⑴ 怎麼實現多組數據的同時輸入和輸出
定義一個數組變數用來讀入，例如a[]
i=0;
a[0]=0;
while(a<>-1)do begin
i=i+1;
read(a[i]);
end;
如此輸入數據，輸出也是，用一個數組來存放輸出數據並依次按格式輸出
⑵ C++怎麼輸入多組數據
C++中實現多組數據輸入主要有兩種方式：
1.首先輸入一個n，表示將有n個輸入輸出，例如：
#include<iostream>
usingnamespacestd;
intmain()
{
intn,a;
cin>>n;
while(n--){
cin>>a;
cout<<"輸出："<<a<<endl;
}
return0;
}
/*
運行結果：
3
111
輸出：111
222
輸出：222
333
輸出：333
*/
2.使用while(cin>>a){}語句，直達輸入ctrl+z，結束輸入，例如：
#include<iostream>
#include<stdio.h>
usingnamespacestd;
intmain()
{
inta;
while(cin>>a){
cout<<"輸出："<<a<<endl;
}
return0;
}
/*
運行結果：
5
輸出：5
6
輸出：6
8
輸出：8
^Z
*/
⑶ matlab 怎麼輸入多組數據
n=input('輸入數字個數');
for ii=1:n
x(ii)=input('輸入數字：');
end
mean(x)
⑷ C語言如何實現輸入多組數據測試
循環按照格式讀入每組數據即可。
對於輸入多組數據測試的情況，需要約定結束的類型，常用的有兩種：
1 當讀入數據為一組特定值時，結束測試。
比如每組2個整型數據，以空格分隔，當輸入的兩個數均為-1時，結束測試。代碼可以寫作：
inta,b;
while(1)
{
scanf("%d%d",&a,&b);
if(a==-1&&b==-1)break;//退出測試的條件。
//測試代碼。
}
2 當讀到EOF時，結束測試。
同樣讀入兩個整型數據，以空格分隔，當讀到EOF時結束測試。代碼可以寫作：
inta,b;
while(scanf("%d%d",&a,&b)!=EOF)//當出現EOF時，結束測試。
{
//測試代碼。
}
⑸ 怎麼實現多組數據輸入。格式要求如圖
其實OnlineJudge上的多組數據輸入和我們平時要求的多組數據輸入是一樣的
就是處理好第一組數據之後能再處理下一組輸入
而不是將數據一次性全部輸入
如果有case的數量就用它來計數，主要處理操作用放在循環里
如果沒有的話就用scanf()的返回值
如:while(scanf("%d%d",&a,&b)==2)
{
...
}
⑹ ACM里說多組數據怎麼輸入啊
int n,sum=0;
while(scanf("%d",&n)!=EOF)
{
sum+=n;
}
EOF 處理到文件的結束就可以了。
根據題目的要求在合適的位置輸出就可以了，如果每組數據都要輸出，則寫在循環裡面，如果所有數據加完了在輸出，那麼就在循環外邊。
只要你知道是要處理到文件結束就根據題目的要求做就可以了。
如果sum是求所有的輸入的和，就是放在外邊的！請樓上的兄弟認真思考後再發言。
⑺ 想問一下怎麼輸入多組數據啊，就是需要輸入多組二維數組
inta[5][5];
inti,j,N;
scanf("%d",&N);
while(N--)
{
for(i=0;i<5;i++)
for(j=0;j<5;j++)
scanf("%d",&a[i][j]);
//以上就完成了數組的讀入
//自己補充，查找最大數和四個最小數吧
//以下完成輸出
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
printf("%d",a[i][j]);
printf(" ");
}
printf(" ");
}
⑻ 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
0
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;
}
⑼ c語言如何實現輸入數據有多組
#include <stdio.h>
void main()
{
int a,b,n;
scanf("%d",&n);
while(n<1||n>10)
{
printf("Error!!");
scanf("%d",&n);
}
while (n--)
{
scanf("%d%d",&a,&b);
printf("%d\n",a+b);
}
}
//*************************************************
#include <stdio.h>
void main()
{
int a,b;
while (scanf("%d%d",&a,&b)!=EOF)//以0結束就把0替換EOF
{
printf("%d\n",a+b);
}
}
⑽ c語言如何輸入多組數據
#include<stdio.h>
intmain()
{
intn;
scanf("%d",&n);
while(n!=0)//等於0就退出，不等於0就繼續輸入
{
scanf("%d",&n)；
printf("%d ",n);
}
}