2012 the computer Research of Harbin Institute of Technology The real problem

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Problem Solving Ideas:

Simulation

Full code:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <set >using namespace Std;const int INF = 1000000000;const int maxn = 10001; Char A[MAXN], b[maxn];void mystrcat (char dststr[], char srcstr[]) {    char c[maxn];    int cnt = 0;    for (int i = 0; dststr[i]; i + +)        c[cnt++] = dststr[i];    for (int i = 0; srcstr[i]; i + +)        c[cnt++] = srcstr[i];    c[cnt++] = ' + ';    for (int i = 0; c[i]; i + +)        printf ("%c", C[i]);    cout << Endl;} int main () {    #ifdef doubleq    freopen ("In.txt", "R", stdin);    #endif//Doubleq    while (Cin >> a >> b)    {         Mystrcat (A, b);}    }

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Problem Solving Ideas:

This problem can add difficulty, and feel test instructions expression problem, strangers to the rich talk about the plan, the first tone should be a stranger said .... But in the end is a rich tone to solve the problem.

Full code:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <set >using namespace Std;const int INF = 1000000000;const int maxn = 10001; int main () {    long long sum = 1;    A long long k = 1;    for (int i = 2; I <=; i + +)    {        k = k * 2;        sum + = k;    }    cout << "$" << sum << Endl;}

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Problem Solving Ideas:

This problem let me understand the nine degree above to use the loop to determine whether to enter the end to terminate!!!

Full code:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <set >using namespace Std;const int INF = 1000000000;const int maxn = 10001;    Long Long a[4][4], B[4][4];long long C[4][4];int main () {#ifdef Doubleq freopen ("In.txt", "R", stdin); #endif//Doubleq while (CIN >> a[0][0]) {cin >> a[0][1] >> a[0][2] >> a[1][0] &        Gt;> a[1][1] >> a[1][2]; for (int i = 0; i < 3; i + +) {for (int j = 0; J < 2; J + +) {cin >            > B[i][j];                }} for (int i = 0; i < 2; i + +) {for (int j = 0; J < 2; J + +) {                A long long sum = 0;                for (int k = 0; k < 3; k + +) {sum + = a[i][k] * B[k][j];            } c[i][j] = sum;       }} for (int i = 0; i < 2; i + +) {     for (int j = 0; J < 2; J + +) {cout << c[i][j] << "";        } cout << Endl;  }    }}

2012 the computer Research of Harbin Institute of Technology The real problem