金华网站制作案例,网站建立快捷方式,网站设计师岗位职责,做网站的核验单 是下载的吗1#xff1a;置换的轮换表示 给出一个置换#xff0c;写出该置换的轮换表示。比如 (1 2 3 4 5 6 7 8 9) (3 1 6 2 9 7 8 4 5) 表示为(1 3 6 7 8 4 2)(5 9) 输入#xff1a; 置换后的序列 输出#xff1a; 不相杂的轮换乘积#xff0c;每行表示一个轮换#xff08;轮换的起… 1置换的轮换表示 给出一个置换写出该置换的轮换表示。比如 (1 2 3 4 5 6 7 8 9) (3 1 6 2 9 7 8 4 5) 表示为(1 3 6 7 8 4 2)(5 9) 输入 置换后的序列 输出 不相杂的轮换乘积每行表示一个轮换轮换的起始数字最小每个轮换的起始数字递增排序单轮换省略 例如 样例1 输入空格分隔 3 1 6 2 9 7 8 4 5 输出 1 3 6 7 8 4 2空格分隔 5 9空格分隔 样例2 输入空格分隔 4 6 7 5 1 2 3 输出 1 4 5空格分隔 2 6空格分隔 3 7 空格分隔 #include cstdio
#include cstring
int main()
{char c[50];gets(c); int n(strlen(c)1)/2;int i,j0,k,flag1,a[25],b[25];for(i0;in;i){a[i]c[2*i]-48;b[i]0;}int l1;b[j]l;printf(%d ,b[j]);j;for(i0;jn;){if(a[i]l) //a[i]等于这一轮换的头说明轮换结束 {printf(\n); l; //找下一个轮换的头while(flag1){for(k0;kj;k){ if(lb[k]) //l不能已经在之前的轮换中 {break;}}if(kj) //l不在轮换里 {if(la[l-1]) //单轮换不输出还要接着找头不改变flag跳出while循环{b[j]l;j; if(jn) //如果单轮换是最后的数就不用再找了跳出while循环 {flag0;}} else //找到l了可以跳出while循环{flag0; b[j]l;printf(%d ,b[j]);j;il-1;} }else{l;}}flag1; //给下一次找头做准备 }else //a[i]不等于轮换的头 就继续加到轮换里{b[j]a[i];printf(%d ,b[j]);j;ia[i]-1;}}return 0;
} 2轮换的复合运算 集合S中的元素个数小于10个,输入该集合的两个轮换t1和t2计算两个轮换的乘积t乘积采用左复合输出的乘积结果t也表示为轮换。输入时每行输入一个不相交轮换tii1,2,输出的乘积为不相交轮换单轮换省略不写。 样例1 输入 (注数字和括号之间都没有空格,括号“(”和)为半角英文) (123)(456) 注t1 (245) 注t2 输出 (1253)(46) 注乘积t的每个轮换的起始数字是递增排序 样例2 输入 (注数字和括号之间都没有空格,括号“(”和)为半角英文) (16)(23) (45) (36)(125) 输出 (13)(2456) 注每个轮换的起始数字是递增排序 样例3 输入 (注数字和括号之间都没有空格,括号“(”和)为半角英文) (2654) (15) 输出 (14265) 注每个轮换的起始数字是递增排序 样例4 输入 (注数字和括号之间都没有空格,括号“(”和)为半角英文) (1392)(475) (267) 输出 (13926547) 注每个轮换的起始数字是递增排序 #include iostream
#include cstring
using namespace std;
int b[20];
void work(int a[],int n){int j,i0,k0,num1,countn-3;printf((1);int b[10]{0};while(kcount-1){if(a[i]!i1)printf(%d,a[i]);b[i]1;k;ia[i]-1;if(a[i]num){b[i]1;k;for(j0;jcount-1;j){if(b[j]0){ij;numj1;if(num!a[i])printf()(%d,num);break;}}}}printf());
}int main()
{char x[20],y[20];gets(x);gets(y);int i,j,k;int num10,num20,e0,r0;int count0;int z[20]{0};for(i1;i11;i) //从1开始 {for(j0;jstrlen(y);j) //处理下面的轮换 {if(y[j]-0 i y[j1]!)) //不等于末尾数 {num1y[j1]-0;break;}if(y[j]-0i y[j1])) //等于末尾数 {int m0;for(mj;m0;m--){ if(y[m]() //找到该轮换的第一个数 {num1y[m1]-0;e1;break;}}if(e1){break;}}num1i; //未找到这个数即等于自身。 }for(k0;kstrlen(x);k){if(x[k]-0num1 x[k1]!)) //不等于末位数 {num2x[k1]-0;break;};if(x[k]-0num1 x[k1])){int m0; for(mk;m0;m--){if(x[m](){num2x[m1]-0;r1;break;}}if(r1){break;}};num2num1;}e0;r0;z[count]num2; count;}work(z,count);
}
/*
int main(){char c[50];gets(c);char b[50];gets(b); int i,j,k,num10,num20,e0,r0,count0;int arr[50]{0};for(i1;i11;i){//start from 1for(j0;jstrlen(b);j){if(b[j]-0i b[i1]!)){//非末尾的映射num1b[j1]-0;break; } else if(b[j]-0i b[i1])){//末尾的映射int t;for(tj;t0;t--){if(b[t](){//轮换的第一个数num1b[t1]-0;e1;break; }}if(e1){break;}}num1i;//not found}for(k0;kstrlen(c);k){if(c[k]-0num1 c[k1]!)){num2c[k1]-0;break;}else if(c[k]-0num1 c[k1])){int t0;for(tk;t0;t--){if(c[t](){num2c[t1]-0;r1;break;}}if(r1){break;}}num2num1;}e0;r0;arr[count]num2;}//arr [] , countcout((1);int ret[10]{0};i0;j0;k0;int ncount-3,num1;while(kn-1){if(arr[i]!i1){coutarr[i];}ret[i]1;k;iarr[i]-1;if(arr[i]num){ret[i]1;k;for(j0;jn-1;j){if(ret[j]0){ij;numj1;if(num!arr[i]){cout)(num;break;}}}}}cout);return 0;
}
*/
好像给的案例最多只有两个划分区域
// int lenstrlen(c),heap[20],flag0,cnt0;
// for(int i0;ilen;i){
// if(c[i](){
// flag1;
// continue;
// }
// else if(c[i])){
// flag0;
// heap[cnt]-1;
// //以-1分割
// }
// //利用flag判断是否进数组
// if(flag1){
// heap[cnt]c[i]-0;
// }
// }
//
// int lenbstrlen(b),heapb[20],flagb0,cntb0;
// for(int i0;ilenb;i){
// if(b[i](){
// flagb1;
// continue;
// }
// else if(b[i])){
// flagb0;
// heapb[cntb]-1;
// //以-1分割
// }
// //利用flag判断是否进数组
// if(flagb1){
// heapb[cntb]b[i]-0;
// }
// }
// for(int i0;icntb;i){coutheapb[i] ;}
//
// int max-1;
// for(int i0;icnt;i){
// if(maxheap[i]){
// maxheap[i];
// }
// }
// for(int i0;icntb;i){
// if(maxheapb[i]){
// maxheapb[i];
// }
// }
// //coutmax;
//
// //开始映射
// int visit[max1]{0};//标记是否visited
// visit[0]1;
// while(1){
//
// } 3脾气牛排队 Cow Sorting Description Farmer Johns N (1 ≤ N ≤ 10,000) cows are lined up to be milked in the evening. Each cow has a unique grumpiness level in the range 1...100,000. Since grumpy cows are more likely to damage FJs milking equipment, FJ would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (not necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes FJ a total of XY units of time to exchange two cows whose grumpiness levels are X and Y. Please help FJ calculate the minimal time required to reorder the cows and give the procedure of sorting. 输入 第1行N (整数N代表牛的数量) 第2行到N1行牛的脾气值 输出 第1行最小排队代价值 第2行给牛按脾气值升序排列的排队过程每行表示一对交换脾气值空格分隔 样例 输入 7 6 7 4 5 3 2 1 输出 33 6 1 4 5 3 2 7 6 2 4 5 3 1 7 1 2 4 5 3 6 7 1 2 4 3 5 6 7 1 2 3 4 5 6 7 #include cstdio
int main()
{int i,j0,k1,flag1,m,n,sum0;int a[20],b[20],c[20][20];scanf(%d,n); for(i0;in;i){scanf(%d,a[i]);}int t0,l1;b[j] l;j;c[t][k] l;k;for(i0;jn;){if(a[i]l || jn) //a[i]等于这一轮换的头或者b[]满了所有数都进入了轮换说明轮换结束{c[t][0] k; //c的每行第一个元素用来记录这个轮换的阶数1if(c[t][1]*(k-2) c[t][1]*2k) //如果该轮换开头的最小的数仍然很大设为big用big做k-2次对换的代价是big*(k-2)先把1和big进行一次对换用1去做k-2次兑换后再换回来代价是(1big)*21*(k-2){int big c[t][1];for(m0;mk;m){ c[t1][m] c[t][m]; //把当前轮换往下挪一步}c[t][0] 3;c[t][1] 1;c[t][2] c[t1][1]; //加入一行轮换对换 把1和开头换一次c[t1][1] 1; //把1换成该轮换的开头 sum 1big;t;for(k--;k1;k--){ sum c[t][1]c[t][k]; //代价加上对换的两个数}t; c[t][0] 3;c[t][1] big;c[t][2] 1; //再加入一行轮换对换 把1和开头换回来sum 1big;}else{for(k--;k1;k--) //每个轮换都会被拆成k-1个对换{ sum c[t][1]c[t][k]; //代价加上对换的两个数} } t;l;while(flag1){for(m0;mj;m){ if(lb[m]){break; //l不能已经在之前的轮换中 }}if(mj) //l不在轮换里 {if(la[l-1]){ b[j] l;j; //单轮换不输出还要接着找头不改变flag跳出while循环 if(jn){flag 0; //如果单轮换是最后的数就不用再找了跳出while循环 } } else{flag 0; b[j] l;j;c[t][k] l;k;i l-1;} }else{l;}}flag1; //给下一次找头做准备}else{ b[j] a[i];j; //b记录出现在所有轮换里的数 c[t][k] a[i];k; //c记录每一次轮换里的数 i a[i]-1;}} printf(%d\n,sum);jt;for(t0;tj;t) //把每个轮换c[t]拆成c[t][0]-1个对换{ for(kc[t][0]-1;k1;k--) //对换c[t][1]和c[t][k]所代表的数在a[]中的位置{int x,y; //x表示c[t][1]代表的数的位置y表示c[t][k]代表的数的位置for(x0;xn;x){if(a[x]c[t][1]){break;}}for(y0;yn;y){if(a[y]c[t][k]){break;}}int t a[y];a[y] a[x];a[x] t;for(m0;mn;m) //输出此次对换结果{printf(%d ,a[m]);} printf(\n);}} return 0;
} 4洗牌机问题 Description Alice and Bob have a set of N cards labelled with numbers 1 ... N (so that no two cards have the same label) and a shuffle machine. We assume that N is an odd integer. The shuffle machine accepts the set of cards arranged in an arbitrary order and performs the following operation of double shuffle : for all positions i, 1 i N, if the card at the position i is j and the card at the position j is k, then after the completion of the operation of double shuffle, position i will hold the card k. Alice and Bob play a game. Alice first writes down all the numbers from 1 to N in some random order: a1, a2, ..., aN. Then she arranges the cards so that the position ai holds the card numbered ai1, for every 1 i N-1, while the position aN holds the card numbered a1. This way, cards are put in some order x1, x2, ..., xN, where xi is the card at the ith position. Now she sequentially performs S double shuffles using the shuffle machine described above. After that, the cards are arranged in some final order p1, p2, ..., pN which Alice reveals to Bob, together with the number S. Bobs task is to guess the order x1, x2, ..., xN in which Alice originally put the cards just before giving them to the shuffle machine. Input The first line of the input contains two integers separated by a single blank character : the odd integer N, 1 N 1000, the number of cards, and the integer S, 1 S 1000, the number of double shuffle operations. The following N lines describe the final order of cards after all the double shuffles have been performed such that for each i, 1 i N, the (i1)st line of the input file contains pi (the card at the position i after all double shuffles). Output The output should contain N lines which describe the order of cards just before they were given to the shuffle machine. For each i, 1 i N, the ith line of the output file should contain xi (the card at the position i before the double shuffles). 样例1 输入 7 4 6 3 1 2 4 7 5 输出 4 7 5 6 1 2 3 #includeiostream
#includecstdio
#includecstring
#includealgorithm
#includecmath
#define N 1003
using namespace std; int n,m;
int cnt[N],a[N],ans[N],l[N],use[N],b[N]; int main(){ scanf(%d%d,n,m); int maxn0;int j,i;for(i1;in;i){scanf(%d,a[i]),cnt[i]a[i];} bool flag1; int t0;while(1){t;for(i1;in;i){b[i]a[a[i]];}for(i1;in;i){if(cnt[i]!b[i]) {flagfalse;break;}}if(flag){break;}flagtrue;for(int i1;in;i){a[i]b[i];}}for(i1;in;i){a[i]b[i];}tt-m%t;for(i1;it;i){for(j1;jn;j){b[j]a[a[j]];}for(j1;jn;j){a[j]b[j];}}for(i1;in;i){printf(%d\n,a[i]);}return 0;
}
