1、 break; if(pstri0!=kaitou_x) if(*(pjielong-1)= pstri0) pjielong-;return 0;运行结果:二、方格问题main()int i,n,g,j;int f1101=0,f2101=0;int f3101;scanf(f1100=1;f2100=3;for(i=3;=n;i+)g=0;for(j=100;j=1;j-)g=g+2*f1j+f2j;f3j=g%10;g=g/10;for(j=1;j=100;j+)f1j=f2j;f2j=f3j;j=1;while(f3j=0)j+;for(i=j;printf(,f3i);n运行结果二、
2、 FBZ串 int i,j,k,n,temp; int s10002; char q2000=0; temp=1;k); if(k=1) n=1; else for(i=2;=k; n=temp*2;temp=n; =n-1; for(j=0; scanf(sij); if(sij=0) qn*2+i*2+j=Z; else qn*2+i*2+j=B if(si0=si1)&(si0=0) qn+i= else if(si0=si1)&(si0=1) else F for(j=n;=2;j=j/2) for(i=0;=j-1;i=i+2) if(qj+i=qj+i+1)&(qj+i!=) q(
3、j+i)/2=qj+i; else q(j+i)/2=,q1); for(i=1;=n*2-1;i=i+) if(qi=) printf(%c%c,qi*2,qi*2+1); getch(); 四、 分油main () int q1004=0,0,0,0,10,0,0,0; int j,front,rear,x10,x7,x3,y10,y7,y3,p,p1,i; front=1;rear=1;p=1; while(p) x10=qfront0; x7=qfront1; x3=qfront2; front+; if(x10=5)&(x7=5) p=0; else if(x77) y10=x10+
4、x7-7;y7=7;y3=x3; p1=1; for (i=1;=rear; if (y10=qi0)&(y7=qi1) p1=0; if (p1=1) rear+; qrear0=y10; qrear1=y7; qrear2=y3; qrear3=front-1; if (x30) y10=x10+x7; y7=0; y3=x3; rear+; if (x70)&(x3 y7=x7+x3-3;y3=3; else y7=0;y3=x7+x3; if (x3 y10=x10+x3; y3=0; if (x70) y7=7;y3=x7+x3-7; else y7=x7+x3;y3=0; fron
5、t-; for(j=1;=front; printf (%d %d %d n,qj0,qj1,qj2);五、 关键点和桥#include stdio.h int g1010=0,0,0,0,0,0,0,0,0,0, 0,0,1,1,1,0,0,0,0,0, 0,1,0,1,0,0,1,0,0,0, 0,1,1,0,1,1,0,0,0,0, 0,1,0,1,0,1,0,0,0,0, 0,0,0,1,1,0,1,0,0,0, 0,0,1,0,0,1,0,1,0,0, 0,0,0,0,0,0,1,0,1,0, 0,0,0,0,0,0,0,1,0,1, 0,0,0,0,0,0,0,0,1,0; int
6、 g01010=0; int i,j,x,front,rear,s,n; int a,b; int flag10=0; int q10; n=9;关键点为: for(i=5;i+)/关键点检索 g01010=g1010; g0ii+1=0; g0i+1i=0; flagi=1; q1=i; front=1; rear=1; while(front=rear) x=qfront;front+; for(j=1; if(flagj=0)&(gxj=1) rear+;qrear=j;flagj=1; if(rear=n-1) printf( %d,i);/广度优先循环搜索判断关键点,不符合条件记录输
7、出。 printf(桥为: for(a=6;aa+)/桥检索 g01010=g1010; g0aa+1=0; g0a+1a=0; flaga=1; q1=a; front=1; rear=1; while(front for(b=1;bb+) if(flagb=0)&(gxb=1)qrear=b;flagb=1; n-1) %d%d,a,a+1); getch();关键点为: 6 7 8 67 78 89六、 矩阵 int i,j,n,k,n1,s,s0,s1,x,y,p; int g2020; int b20; k=0; k+; for(j=i;n-i; gij=k; gji=k; for(
8、j=n-i-1;i; gn-i-1j=k; gjn-i-1=k; for(j=0;j+) printf(%d ,gij); s0=0; n1=n+n-2;i+) bi=0; for(i=n;n+n-1;i+) bi=1; while(b0=0) s1=1;x=1;y=1;=n1; if(bi=0) x+; else y+; s1=s1+gxy; p=1; j=2; while(p)&(js1) if(s1%j=0) p=0; else j+; if(p) s0+; if ( bs1) s0+; j=n1; while(bj=0) j-; s1=0; while(bj=1) s1+; bj=0;
9、j-; bj=1; s1-; k=n;=s1; bk=1; k-;,s0);七、骑士问题 int i,k,m,n,top,x,y,w,q,x0,y0; int dx9=0,1,2,2,1,-1,-2,-2,-1; int dy9=0,-2,-1,1,2,2,1,-1,-2; int stack1000; int g100100=0;%d,%d,%dn,&x0,&y0); while(n%2=1)&(x0+y0)%2=1) printf(input error gx0y0=1; x=x0; y=y0; top=0; while(top9) k=stacktop; top-; gxy=0; x=x
10、-dxk; y=y-dyk; w=x+dxk; q=y+dyk; if(w=1)&(w(q(gwq=0) x=w; y=q; top+; gxy=1; stacktop=k; k=0;%d %d,x,y);=top; x=x+dxstacki; y=y+dystacki;-%d %d八、士兵排队int x,y,i,i1,j,k,s,s1,n,p,j1,j2; int q100; int g77=0,0,0,0,0,0,0, 0,0,0,0,0,0,1, 0,0,0,0,1,1,0, 0,0,0,0,0,0,0, 0,1,0,0,0,1,0, 0,1,0,1,0,0,1, 0,0,0,1,0,0
11、,0; n=6; k=9;j1=1;(j11)error2 else qj1=j2; j1+;i+) gij2=1;i+) gj2i=0; if(p) for(i=1;,qi);九、HUFFMAN树编码stdlib.hint main() int i,i0,i1,j,j1,j0,n,n1,n2,cmin1,cmin,s,k,p; char str30=wabcddeacefgfg; int tree5213=0; int h27=0; int d27; int code2721; str0=13;p);=str0; hstri-96+; n=0;=26; if(hi n+; treen1=hi
12、; dn=i; if(n0) n1=n+(p-1)-n2; s=1+(n1-p)/(p-1); n2=n1;=s; n2+;cmin1=0;=p; cmin=1000; for(k=1;kn2;k+) if(treek0=0)&(treek1cmin) j1=k;cmin=treek1; cmin1=cmin1+cmin; treen22+p=j1; treej12=n2; treej10=1; treen21=cmin1; for(k=0;=20; codeik=-1; i0=i;i1=treei2;k=21; while(i1!=0) k-; for(j1=3;j1cmax) j1=j;
13、cmax=gij; gij=0;%dn,cmax); for(i=j1-cmax+1;=j1;,ai);Efghi十一、迷宫问题#define M 5 /行数#define N 7 /列数 int k,top,x,y,w,q,p,i; int dx9=0,1,1,1,0,-1,-1,-1,0; int dy9=0,-1,0,1,1,1,0,-1,-1;/ char b100;/ char a100; char stack100;/ 栈的最多储存元素int flag2020=0;int count;/计数 int gM+1N+1= 0,0,1,1,1,1,0,0, 1,0,1,1,0,0,0,1, 1,0,0,1,1,1,0,1, 0,0,1,1,1,1,0,1, 0,1,0,1,1,1,0,0, 1,1,1,0,0,0,0,1; g00=1;/从第1行第1列开始k=0; x=0;y=0; while( p) k+; if(k8)/八个方向 k=stacktop;top-;gxy=0; x=x-dxk; y=y-dyk; w=x+dxk
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