for i:=n downto 1

for i:=n downto 1 数据结构语句FOR i:=n DOWNTO 1 该语句共执行频度为几? 每个语句的频度是多少啊for i:=n downto 1 do (1)beginx:=x+1; (2)for j:=n downto i do (3)y:=y+1; 4)end; 求教关于for i:=n downto 1 如题,要计算该语句的执行频度,答案是n+1,原来程序是这样的：FOR i:=n DOWNTO 1 DOBEGINX:=X+1;FOR J:=n DOWNTO i DOy:=y+1;END; program Noname0; const n=7; var a:array[1..n] of integer; i,j,k,t:integer; begin write('ggg'); for program Noname0;const n=7;var a:array[1..n] of integer;i,j,k,t:integer;beginwrite('ggg');for i:=1 to n do read(a[i]);for i:=1 to n-1 dofor j:=n downto pascal高精度计算1!+2!+…+n!求改用高精度计算出S=1!+2!+3!+...+n!(n=10 then begin h[q]:=h[q]-10;h[q+1]:=h[q+1]+1;end;end;end;i:=100;while h[i]=0 do i:=i-1;for j:=i downto 1 do write(h[i]);end.--------------------------------------分割 pascal 0/1背包和完全背包的差别?0/1背包?for i:=1 to n dofor j:=m downto w[i] do完全背包?for i:=1 to n dofor j:=w[i] to m do两个什么差别?怎么体现?有没有样例可以体现两个的差别?就是输入一样,输出不一样. 第K极值 pascal 求问错在哪了我的程序：var a:array[1..10000] of integer;n,k,i,j,m,t:integer;l,q:boolean;beginreadln(n,k);for i:=1 to n do read(a[i]);for i:=n downto 1 dofor j:=1 to i-1 dobegin if a[j] >a[i] then begint:=a[i];a[i]:=a[j];a[ pascal语言输出下列图形的问题,大家看看我为什么错了?编程输出下列图形 var i,j:integer;beginfor i:=1 to 9 dobegin write ('1':40-3*i);for j:=2 to i dobeginwrite (j:3);end;for j:=i-1 downto 1 beginwrite (j:3);end;writeln ;write pascal题目 求详解var i,j,k,s:integer;begins:=0;for i:=3 downto 1 dobeginfor j:=1 to 3 dobegink:=0;repeatk:=k+1;s:=s+k;until k=j;end;s:=s-(k+1);end;write('s=',s);end. pascal For i:=1 to n 求从头到尾解释一个程序 program ee（input,output); var i,j,k,s:=integer; begin s:=0;for i:=3 downto 1 dobeginfor j:=1 to 3 dobegink:=0;repeatk:=k+1;s:=s+k;until k=j;end;s:=s-(k+1);end;write('s=',s);end.为什么s=18以及for语句的阅读 for(i=0;n vhdl if(x=1) then y'0'); 其中y:buffer std_logic_vector(3 downto 0) VHDL中'1'& f(17 downto 9) & f(8 downto 0) & 关于lingo程序的,sets:A/1..4/:c,b,n;X/A,A/:D;endsetsmin=1.1*n(1)+1.2*n(2)+1.3*n(3)+1.4*n(4);@for(A(i):@sum(A(j):c(i)*D(i,j)=1750);@for(A(i):@sum(A(j):n(i)*D(i,j)=b(i));@for(A(i):@gin(n(i)));@for(A(j):@for(A(i):@gin(D(i,j))));n(1)>n(2);n(2)>n(3);n main() {long m,n,i=1,j,s; scanf(%ld,%ld,&m,&n); for(;i for(i=1; i7; j--) { putchar('$'); } putchar(' '); }