首页 > 代码库 > NYOJ 233 Sort it【冒泡排序】
NYOJ 233 Sort it【冒泡排序】
求解交换次数,用冒泡刚好
Sort it
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描述
- You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.- 输入
- The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.
- 输出
- For each case, output the minimum times need to sort it in ascending order on a single line.
- 样例输入
3 1 2 3 4 4 3 2 1
- 样例输出
0 6
- 来源
- ZJFC 2009-3 Programming Contest
- 上传者
- 张洁烽
#include<stdio.h> #include<string.h> int arr[1100]; int main() { int n,i,j,sum,t; while(~scanf("%d",&n)) { sum=0; memset(arr,0,sizeof(arr)); for(i=0;i<n;i++) scanf("%d",&arr[i]); for(i=1;i<=n-1;i++) for(j=0;j<=n-i-1;j++) if(arr[j]>arr[j+1]) { t=arr[j]; arr[j]=arr[j+1]; arr[j+1]=t; sum++; } printf("%d\n",sum); } return 0; }
NYOJ 233 Sort it【冒泡排序】
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