nyoj 可以直接写的简单题
Triangular Sums nyoj 122
#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
for(int count=1;count<=N;count++)
{
int m, num=0;
cin>>m;
for(int i=1;i<=m;i++)
num += i*((i+1)*(i+2)/2);
cout<<count<<" "<<m<<" "<<num<<endl;
}
return 0;
}
爱摘苹果的小明 nyoj 50
#include<iostream>
using namespace std;
int main()
{
int N;
int a[10];
cin>>N;
while(N--)
{
for(int i=0;i<10;i++)
cin>>a[i];
int m, count=0;
cin>>m;
for(int i=0;i<10;i++)
if(a[i] <= (m+30))
count++;
cout<<count<<endl;
}
return 0;
}
国王的魔镜 nyoi 264
我用到了字符串的逆序
#include<iostream>
#include<string>
using namespace std;
int main()
{
int N;
string str;
cin>>N;
while(N--)
{
cin>>str;
if((str.size() % 2) == 1)
{
cout<<str.size()<<endl;
continue;
}
else
{
string s(str);
while(str.size()%2 == 0 && str == s)
{
s = str.substr(str.size()/2, str.size()/2);
string s1(s.rbegin(),s.rend());
s = s1;
str = str.substr(0, str.size()/2);
}
if(str.size() %2 == 1 && str == s)
cout<<str.size()<<endl;
else
cout<<str.size()+s.size()<<endl;
}
}
return 0;
}
把两个串合并成一个，然后逆序
#include<iostream>
#include<string>
using namespace std;
int main()
{
string str;
string str2;
int N;
cin>>N;
while(N--)
{
cin>>str>>str2;
str += str2;
string s(str.rbegin(), str.rend());
for(int i=0;i<s.length();i++)
{
if(s.at(i) >= 'a' && s.at(i) <= 'z')
cout<<s.at(i);
}
cout<<endl;
}
return 0;
}
不高兴的小明 nyoj 53
#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
int max=0, a, b, index;
for(int i=1;i<=7;i++)
{
cin>>a>>b;
if(a + b > max)
{
max = a+b;
index = i;
}
}
if(max <=8)
cout<<0<<endl;
else
cout<<index<<endl;
}
return 0;
}
分数拆分 nyoj 66
直接写
#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
int m;
cin>>m;
for(int i=m+1;i<=2*m;i++)
{
if((i* m % (m-i)) == 0)
cout<<"1/"<<m<<"=1/"<<(i*m)/(i-m)<<"+1/"<<i<<endl;
}
}
return 0;
}
另类乘法 nyoj 121
#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
int a, b, m, temp;
cin>>a>>b;
double sum = 0;
while(a)
{
m = a % 10;
a = a / 10;
temp = b;
while(temp)
{
sum += m * (temp%10);
temp /= 10;
}
}
cout<<sum<<endl;
}
return 0;
}
此题最优程序
#include<iostream>
#include<string>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
string s1, s2;
cin>>s1>>s2;
int sum =0;
for(int i=0;i<s1.size();i++)
for(int j=0;j<s2.size();j++)
sum += (s1.at(i)-'0')*(s2.at(j)-'0');
cout<<sum<<endl;
}
return 0;
}
#include<iostream>
#include<string>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
string str;
int m, second=0, div=1;
cin>>m;
while(m--)
{
cin>>str;
if(str == "IN")
div *= 20;
else if(str == "STAY")
{
int stay;
cin>>stay;
second += (stay * 60) / div;
}
else
div /= 20;
}
cout<<second<<endl;
}
return 0;
}
小明求素数积 nyoj 225
#include<iostream>
using namespace std;
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009};
int main()
{
int N, m, num;
cin>>N;
while(N--)
{
cin>>m;
num = 1;
for(int i=0;prime[i]<=m;i++)
{
num *= prime[i];
num %= 1000000;
}
cout<<num<<endl;
}
return 0;
}
玩转矩阵的C小加 nyoj 261
#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
int a[3][3];
cin>>a[0][0]>>a[0][1]>>a[0][2]>>a[1][0]>>a[1][1]>>a[1][2]>>a[2][0]>>a[2][1]>>a[2][2];
cout<<a[0][0]<<" "<<a[1][0]<<" "<<a[2][0]<<endl<<a[0][1]<<" "<<a[1][1]<<" "<<a[2][1]<<endl<<a[0][2]<<" "<<a[1][2]<<" "<<a[2][2]<<endl;
}
return 0;
}
正三角形外接圆面积 nyoj 274
#include <iostream>
using namespace std;
#define PI 3.1415926
int main()
{
int m;
cin>>m;
while(m--)
{
double s,a;
cin>>a;
s=PI*a*a/3.0;
cout.setf(ios_base::fixed);
cout.precision(2);
cout<<s<<endl;
}
return 0;
}
算菜价 nyoj 316
#include<stdio.h>
int main()
{
double a,b,sum=0;
char c[20];
while(scanf("%s%lf%lf",c,&a,&b)!=EOF)
{
sum=sum+a*b;
}
printf("%.1lf\n",sum);
return 0;
}
猴子吃桃 nyoj 324
#include<iostream>
using namespace std;
int main()
{
int N;
cin>>N;
while(N--)
{
int m;
cin>>m;
int num=1;
for(int i=0;i<m;i++)
{
num = (num+1) * 2;
}
cout<<num<<endl;
}
return 0;
}
整除个数 nyoj 399
#include<iostream>
using namespace std;
int main()
{
int n, b;
while(cin>>n>>b)
cout<<n/b<<endl;
return 0;
}
黑色帽子 nyoj 455
#include<iostream>
using namespace std;
int main()
{
unsigned int N, m;
cin>>N;
while(N--)
{
cin>>m;
cout<<m<<endl;
}
return 0;
}
九九乘法表 nyoi 463
#include<iostream>
using namespace std;
int main()
{
int N,M;
cin>>N;
while(N--)
{
cin>>M;
for(int i=1;i<=M;i++)
{
cout<<i<<"*"<<i<<"="<<i*i;
for(int j=i+1;j<=9;j++)
cout<<" "<<i<<"*"<<j<<"="<<i*j;
cout<<endl;
}
}
return 0;
}
A+B problem III
#include<iostream>
using namespace std;
int main()
{
int N;
double a , b, c;
cin>>N;
while(N--)
{
cin>>a>>b>>c;
if(((a+b)-c) < 0.0001 && ((a+b)-c) > -0.0001)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
return 0;
}
日期计算 nyoj 75
#include<iostream>
using namespace std;
bool isLeap(int year)
{
if(((year%4 == 0) && (year%100 != 0)) || year%400 == 0)
return true;
return false;
}
int main()
{
int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int N;
cin>>N;
int year, month, day;
while(N--)
{
int num = 0;
cin>>year>>month>>day;
for(int i=1;i<month;i++)
{
num += a[i];
}
num += day;
if(isLeap(year) && (month >= 3))
num++;
cout<<num<<endl;
}
return 0;
}
成绩转换 nyoj 98
#include<iostream>
using namespace std;
int main()
{
int N, score;
cin>>N;
while(N--)
{
cin>>score;
if(score >= 90 && score <= 100)
cout<<"A"<<endl;
else if(score >= 80 && score <= 89)
cout<<"B"<<endl;
else if(score >= 70 && score <= 79)
cout<<"C"<<endl;
else if(score >= 60 && score <= 69)
cout<<"D"<<endl;
else if(score >= 0 && score <= 59)
cout<<"E"<<endl;
}
return 0;
}