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+ # Dimik - Small to Large
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+
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+ There will be three different numbers. They have to be printed from small to large sizes.
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+
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+ ### Solution
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+ * Case number will be according to the number of test case.
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+ * Finding the middle number which is not min or max is the main task.
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+
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+ ### C++
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+ ``` cpp
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+ #include < iostream>
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+ #include < algorithm>
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+ using namespace std ;
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+
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+ int main (){
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+ int test_case;
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+ cin >> test_case;
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+
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+ for (int i = 1; i <= test_case; i++){
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+ int n1, n2, n3;
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+ cin >> n1 >> n2 >> n3;
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+ int a = min(min(n1, n2), n3);
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+ int b = max(max(n1, n2), n3);
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+ if (n1 != a && n1 != b)
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+ {
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+ cout << "Case " << i <<": " << a << " " << n1 << " " << b << endl;
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+ } else if (n2 != a && n2 != b){
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+ cout << "Case " << i <<": " << a << " " << n2 << " " << b << endl;
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+ } else if (n3 != a && n3 != b){
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+ cout << "Case " << i <<": " << a << " " << n3 << " " << b << endl;
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+ }
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+ }
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+ return 0;
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+ }
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+ ```
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