算法设计与分析 第二版 吕国英 第四章课后习题答案.doc

4.1#include#includevoid main()int i,n,sum=0,power,proce,remainder,interger,decimal;printf(请输入2+22+2222(n个2)(不考虑精度)的nnn=);scanf(%d,&n);for(i=1;i=n&i9)for(i=1,power=n-9;i=5)sum+;interger=sum/100000000;decimal=sum%100000000;printf(2+22+2222(n个2)(不考虑精度)=%d.%d*e+%dn,interger,decimal,n-1);elseprintf(2+22+2222(n个2)(不考虑精度)=%dn,sum);4.2#includevoid main()int i,total=2,day=10;for(i=day-1;i=1;i-)total=(total+2)*2;printf(第一天共有%d个桃。n,total);4.3#includevoid main()int flag=0,last=54,card;printf(共有牌：%d 张n,last);printf(计算机出：3 );last=last-3;while(last1)printf(还有牌：%d 张,last);printf(您出：);scanf(%d,&card);printf(计算机出：%d ,5-card);last=last-5;printf(n还有牌：%d 张。您只能输入1。n,last);printf(您出：);scanf(%d,&card);printf(Sorry!您输了n);4.5#includevoid main()int array88,i,j,i1=0,i2=1,i3=2,i4=3,i5=4,i6=5,i7=6,i8=7,j1=0,j2=0,j3=0,j4=0,j5=0,j6=0,j7=0,j8=0;/先判断不在同一行和同一列，然后再做不在对角线printf(此处预先规定，第一行第一列处是皇后。n);for(j2=1;j28;j2+)for(j3=1;j38;j3+)for(j4=1;j48;j4+)for(j5=1;j58;j5+)for(j6=1;j68;j6+)for(j7=1;j78;j7+)for(j8=1;j88;j8+)if(j3!=j2)if(j4!=j2&j4!=j3)if(j5!=j2&j5!=j3&j5!=j4)if(j6!=j2&j6!=j3&j6!=j4&j6!=j5)if(j7!=j2&j7!=j3&j7!=j4&j7!=j5&j7!=j6)if(j8!=j2&j8!=j3&j8!=j4&j8!=j5&j8!=j6&j8!=j7)if(i2-i1+j2-j1)!=0&(i2-i1-j2+j1)!=0)/两点的位置关系！if(i3-i1+j3-j1)!=0&(i3-i1-j3+j1)!=0)if(i3-i2+j3-j2)!=0&(i3-i2-j3+j2)!=0)if(i4-i1+j4-j1)!=0&(i4-i1-j4+j1)!=0)if(i4-i2+j4-j2)!=0&(i4-i2-j4+j2)!=0)if(i4-i3+j4-j3)!=0&(i4-i3-j4+j3)!=0)if(i5-i1+j5-j1)!=0&(i5-i1-j5+j1)!=0)if(i5-i2+j5-j2)!=0&(i5-i2-j5+j2)!=0)if(i5-i3+j5-j3)!=0&(i5-i3-j5+j3)!=0)if(i5-i4+j5-j4)!=0&(i5-i4-j5+j4)!=0)if(i6-i1+j6-j1)!=0&(i6-i1-j6+j1)!=0)if(i6-i2+j6-j2)!=0&(i6-i2-j6+j2)!=0)if(i6-i3+j6-j3)!=0&(i6-i3-j6+j3)!=0)if(i6-i4+j6-j4)!=0&(i6-i4-j6+j4)!=0)if(i6-i5+j6-j5)!=0&(i6-i5-j6+j5)!=0)if(i7-i1+j7-j1)!=0&(i7-i1-j7+j1)!=0)if(i7-i2+j7-j2)!=0&(i7-i2-j7+j2)!=0)if(i7-i3+j7-j3)!=0&(i7-i3-j7+j3)!=0)if(i7-i4+j7-j4)!=0&(i7-i4-j7+j4)!=0)if(i7-i5+j7-j5)!=0&(i7-i5-j7+j5)!=0)if(i7-i6+j7-j6)!=0&(i7-i6-j7+j6)!=0)if(i8-i1+j8-j1)!=0&(i8-i1-j8+j1)!=0)if(i8-i2+j8-j2)!=0&(i8-i2-j8+j2)!=0)if(i8-i3+j8-j3)!=0&(i8-i3-j8+j3)!=0)if(i8-i4+j8-j4)!=0&(i8-i4-j8+j4)!=0)if(i8-i5+j8-j5)!=0&(i8-i5-j8+j5)!=0)if(i8-i6+j8-j6)!=0&(i8-i6-j8+j6)!=0)if(i8-i7+j8-j7)!=0&(i8-i7-j8+j7)!=0)printf(*n);for(i=0;i8;i+)for(j=0;j8;j+)arrayij=0;arrayi1j1=1;arrayi2j2=1;arrayi3j3=1;arrayi4j4=1;arrayi5j5=1;arrayi6j6=1;arrayi7j7=1;arrayi8j8=1;printf(1是皇后所在位置，0为其它位置。n);for(i=0;i8;i+)for(j=0;j8;j+)printf(%d ,arrayij);printf(n);4.6#includevoid main()int goods=100,big=0,middle=0,small=0;for(big=0;big=33;big+)for(middle=0;middle=50;middle+)for(small=0;small=100;small=small+2)if(big+middle+small)=100)if(big*3+middle*2+small/2)=goods)printf(大马：%2d 中马:%2d 小马:%2dn,big,middle,small);4.7#includevoid main()int n,flag=1;for(n=7;flag;n=n+7)if(n%2=1)if(n%3=2)if(n%5=4)if(n%6=5)flag=0;printf(该数为%dn,n-7);4.8#includevoid main()int th,flag=0;for(th=1000;th9999;th+)if(th/1000!=th/100%10)if(th/1000!=th/10%10)if(th/1000!=th%10)if(th/100%10!=th/10%10)if(th/100%10!=th%10)if(th/10%10!=th%10)if(th%(th/1000)*(th/1000)=0)flag=0;if(th/100%10)=0)flag=1;elseif(th%(th/100%10)*(th/100%10)=0)flag=1;if(flag=1)flag=0;if(th/10%10)=0)flag=1;elseif(th%(th/10%10)*(th/10%10)=0)flag=1;if(flag=1)flag=0;if(th%10=0)flag=1;elseif(th%(th%10)*(th%10)=0)flag=1;if(flag=1)printf(%dn,th);4.9#include main() int a10=1,10,5,4,9,3,8,6,7,2; int i,max1=0,max2=0,min1=0,min2=0; for(i=0;i10;i+) if(amax1ai) min1=i; if(max1=0) max2=1; if(min1=0) min2=1; for(i=0;i10;i+) if(i=max1|i=min1) continue; if(amax2ai) min2=i; printf(max1=%dnmax2=%dnmin1=%dnmin2=%dn,amax1,amax2,amin1,amin2); 4.16#includeint fun(int N,int basic,int array)int n1=0,n2=0,n3=0,i;n3=(N%2!=0)?N+basic:0;for(i=0;in2?basic+1:basic+2;