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题目连接:点击打开链接
解题思路:
模拟
完整代码:
#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++] = '\0';
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);
}
}
解题思路:
此题完全可以增加难度的,而且感觉题意表述有问题,陌生人向富翁谈计划,那第一句口吻应该是陌生人说的。。。。可是最后确是以富翁口吻来解题。
完整代码:
#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;
long long k = 1;
for(int i = 2 ; i <= 30 ; i ++)
{
k = k * 2;
sum += k;
}
cout << "300 " << sum << endl;
}
解题思路:
这道题让我明白了九度上面时要用循环判断是否输入结尾才终止!!!
完整代码:
#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] >> 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 ++)
{
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;
}
}
}
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原文地址:http://blog.csdn.net/u013447865/article/details/44887643