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The retired worker resumed his task of servicing the buses without additional pay when he ________at home.
A. could stay
B. could have stayed
C. can stay
D. must have stayed
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更多 “ The retired worker resumed his task of servicing the buses without additional pay when he ________at home. A. could stayB. could have stayedC. can stayD. must have stayed ” 相关考题
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阅读以下说明和C程序,将应填入(n)处的字句写在答题纸的对应栏内。【说明】假设需要将N个任务分配给N个工人同时去完成,每个人都能承担这N个任务,但费用不同。下面的程序用回溯法计算总费用最小的一种工作分配方案,在该方案中,为每个人分配1个不同的任务。程序中,N个任务从0开始依次编号,N个工人也从0开始依次编号,主要的变量说明如下:c[i][j]:将任务i分配给工人j的费用;task[i]:值为0表示任务i未分配,值为j表示任务i分配给工人j;worker[k]:值为0表示工人k未分配任务,值为1表示工人k已分配任务;mincost:最小总费用。【C程序】#include<stdio.h>#define N 8 /*N表示任务数和工人数*/int c[N][N];unsigned int mincost=65535; /*设置min的初始值,大于可能的总费用*/int task[N],temp[N],workerIN];void Plan(int k,unsigned Int cost){ int i;if ((1)cost<mincost){mincost=cost;for (i=0;i<N;i++) temp[i]:task[i];}else{for(i=0;i<N;i++) /*分配任务k*/if (worker[i]=0(2)){worker[i]=1; task[k]=(3);Plan((4),cost+c[k][i]);(5); task[k]=0;}/*if*/}}/*Plan*/void main(){int i,j;for (i=0;i<N;i++) { /*设置每个任务由不同工人承担时的费用及全局数组的初值*/worker[i]=0;task[i]=0; temp[i]=0;for(j=0;j<N;j++)scanf ("%d",c[i][j]);}Plan (0,0); /*从任务0开始分配*/printf("\n最小费用=%d\n",mincost);for(i二0;i<N;i++)pnntf("Task%d iB assigned toWorker%d\n",i,temp[i]);}/*main*/
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