标签:style blog io color os sp for on div
这题第一次看的时候以为是区间替换,后来发现看错了,只是单纯的元素替换。
解题思路:
先对输入的序列加个数组排个序
遍历下来,如果和排序后的结果当前元素不同,设当前位置为 i, 则往下面找,设查找位置为j
使得满足 a[j] == b[i] && a[j] != b[j]
一次遍历即可。易得证
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <cstring> #include <cmath> #include <stack> #include <queue> #include <vector> #include <algorithm> #define ll long long #define Max(a,b) (((a) > (b)) ? (a) : (b)) #define Min(a,b) (((a) < (b)) ? (a) : (b)) #define Abs(x) (((x) > 0) ? (x) : (-(x))) using namespace std; const int INF = 0x3f3f3f3f; int a[3111], b[3111], n, counti; int ans[3111][2]; int main(){ int i, j, t, m; while(EOF != scanf("%d",&n)){ counti = 0; for(i = 1; i <= n; ++i){ scanf("%d",&a[i]); b[i] = a[i]; } sort(b + 1, b + 1 + n); for(i = 1; i <= n; ++i){ if(a[i] != b[i]){ for(j = i + 1; j <= n; ++j){ if(a[j] == b[i] && a[j] != b[j]){ swap(a[j], a[i]); ++counti; ans[counti][0] = j; ans[counti][1] = i; break; } } } } printf("%d\n",counti); for(i = 1; i <= counti; ++i){ printf("%d %d\n",ans[i][1] - 1, ans[i][0] - 1); } } return 0; }
标签:style blog io color os sp for on div
原文地址:http://www.cnblogs.com/wushuaiyi/p/4117973.html