1 #include<
iostream>
2 using namespace std;
3 int main()
4 {
5 int a,b;
6 cin>>a>>
b;
7 cout<<a+
b;
8 return 0;
9 }难点：若用C++注意别漏 using namespace std;
1 #include <stdio.h>
2 int main()
3 {
4 unsigned long long n;
5 scanf(
"%I64u", &
n);
6 printf(
"%I64u\n", n*(n+
1)/
2);
7 return 0;
8 }难点：注意64位用法
1 #include <stdio.h>
2
3 int main()
4 {
5 long long n, sum;
6 scanf(
"%I64d", &
n);
7 sum = (n +
1) * n /
2;
8 printf(
"%I64d", sum);
9 return 0;
10 }难点:注意64位用法
1 #include <stdio.h>
2 #include
<math.h>
3
4 int main()
5 {
6 int r;
7 scanf(
"%d", &
r);
8 printf(
"%.7lf\n",
atan(1.0)*4*r*
r);
9 return 0;
10 }难点： 1.注意输出要求的长度
.7 2.注意PI的计算 atan(1
.0)*4
Fibonacci数列
1 #include <stdio.h>
2 int fun(
int x)
3 {
4 if(
x==1||x==2) //注意递归的结束
5 return 1;
6 else
7 return fun(x-1)+
fun(x-2);
8 }
9 int main()
10 {
11 int n,i;
12 scanf(
"%d",&
n);
13 printf(
"%d",fun(n)
007);
14 return 0;
15 }难点：递归的使用
时间转换 1 #include<iostream>
2 using namespace std;
3 int main()
4 {
5 int t;
6 cin>>
t;
7 int h,m,s;
8 h=t/3600;
9 m=t600/60;
10 s=t600`;
11 cout<<h<<
":"<<m<<
":"<<s<<
endl;
12 return 0;
13 }
字符串小写转大写#include <ctype.h> http://baike.baidu.com/link?url=WH94T6MdbBKwTAqcoGmMz4neQkY7m1odU14d1TxTDm3fHEyhJeGtvJJ9GyX9ZWn3JsfYW1m8Aavez0zfETYjx_1 for(i =
0; i <
sizeof(s); i++
)
2 s[i] =
toupper(s[i]);
矩阵相交面积计算，方法不错 1 #include <iostream>
2 #include
<algorithm> //用来使用max min
3 #include <cmath>
4 #include <cstdio>
5 using namespace std;
6 int main()
7 {
8 double x1, x2, y1, y2;
9 double q1, q2, w1, w2;
10 while (cin >> x1 >> y1 >> x2 >> y2 >> q1 >> w1 >> q2 >>
w2)
11 {
12 double xx =
max(min(x1, x2), min(q1, q2));
13 double yy =
max(min(y1, y2), min(w1, w2));
14 double xxup =
min(max(x1, x2), max(q1, q2));
15 double yyup =
min(max(y1, y2), max(w1, w2));
16 if (xxup >
xx)
17 printf(
"%.2f\n",
fabs((xx)-(xxup))*
fabs((yy)-
(yyup)));
18 else printf(
"0.00\n"); //别忘了不想交的时候输出0
19 }
20 }
转载于:https://www.cnblogs.com/wwjyt/p/3608904.html
相关资源：BS部分蓝桥题库.docx
