本题要求将给定的N个正整数按非递增的顺序，填入“螺旋矩阵”。所谓“螺旋矩阵”，是指从左上角第1个格子开始，按顺时针螺旋方向填充。要求矩阵的规模为m行n列，满足条件：m*n等于N；m>=n；且m-n取所有可能值中的最小值。

输入在第1行中给出一个正整数N，第2行给出N个待填充的正整数。所有数字不超过10⁴，相邻数字以空格分隔。

输出螺旋矩阵。每行n个数字，共m行。相邻数字以1个空格分隔，行末不得有多余空格。

#include <stdio.h>
#include <algorithm>
int cmp(int a,int b)
{return a>b;
}
using namespace std;
int a[10010],c[1001][1001];
int main()
{int N,m,n,i;scanf("%d",&N);for(i=0; i<N; i++)scanf("%d",&a[i]);if(N==1){printf("%d\n",a[0]);}else{int f=200000000;for(i=1; i*i<=N; i++)if(N/i*i==N&&f>N/i-i){f=N/i-i;n=i;m=N/i;}sort(a,a+N,cmp);int j=1,now=0,U=1,D=m,L=1,R=n;i=1;while(now<N){while(j<R&&now<N){c[i][j]=a[now++];j++;}while(i<D&&now<N){c[i][j]=a[now++];i++;}while(j>L&&now<N){c[i][j]=a[now++];j--;}while(i>U&&now<N){c[i][j]=a[now++];i--;}i++;j++;U++,D--;L++,R--;if(now==N-1)c[i][j]=a[now++];}for(i=1; i<=m; i++){printf("%d",c[i][1]);for(j=2; j<=n; j++)printf(" %d",c[i][j]);printf("\n");}}return 0;
}

#include <stdio.h>
#include<string.h>
int main()
{char s[20],*b="parallel";int t,n,i,r;double a[300],sum;scanf("%d",&t);while(t--){scanf("%s",s);scanf("%d",&n);for(i=0; i<n; i++)scanf("%lf%%",&a[i]);r=strcmp(s,b);if(r==0){sum=1.0;for(i=0; i<n; i++)sum=sum*(100-a[i]);for(i=1; i<n; i++)sum=sum*0.01;printf("%.2f%%\n",(100-sum));}else{sum=1.0;for(i=0; i<n; i++)sum=sum*a[i];for(i=1; i<n; i++)sum=sum*0.01;printf("%.2f%%\n",sum);}}return 0;
}

Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

给你n根火柴棍，你可以拼出多少个形如“A+B=C”的等式？等式中的A、B、C是用火柴棍拼出的整数（若该数非零，则最高位不能是0）。用火柴棍拼数字0-9的拼法如图所示：

注意：

1. 加号与等号各自需要两根火柴棍

2. 如果A≠B，则A+B=C与B+A=C视为不同的等式（A、B、C>=0）

3. n根火柴棍必须全部用上

共一行，又一个整数n（n<=24）。

共一行，表示能拼成的不同等式的数目。

2个等式为0+1=1和1+0=1。

本题要求计算A/B，其中A是不超过1000位的正整数，B是1位正整数。你需要输出商数Q和余数R，使得A = B * Q + R成立。

输入在1行中依次给出A和B，中间以1空格分隔。

在1行中依次输出Q和R，中间以1空格分隔。

#include<iostream>
#include<string>
using namespace std;
int main(){string str,ans;int n,d = 0; cin>> str>> n;for(int i = 0; i <= str.size()-1; i++){int current = d * 10 + (str[i]-'0');ans+=(current/n+'0');d=current % n;}for(int i=(ans[0] == '0'&&ans.size()!=1)?1:0;i<ans.size();i++)cout << ans[i];cout << " " << d;return 0;
}

#include <stdio.h>
int main()
{int t;scanf("%d",&t);while(t--){char c;int a,b;getchar();scanf("%c",&c);if(c=='&'){scanf("%d%d",&a,&b);printf("%d\n",a&b);}else if(c=='|'){scanf("%d%d",&a,&b);printf("%d\n",a|b);}else if(c=='^'){scanf("%d%d",&a,&b);printf("%d\n",a^b);}else{scanf("%d",&a);printf("%d\n",~a);}}return 0;
}