yangs_s的博客

NUM.ONE 普通n2做法：var a,f:array[1..500]of longint;var i,j,k,l,m,n,max:longint;begin read(n); for i:=1 to n do read(a[i]); fillchar(f,sizeof(f),0); f[1]:=1; for i:=1 to n do for j:=1 t

NUM.ONE——欧几里得算法（辗转相除法求最大公约数）（循环型）var a,b,i,c,j,k,l,m,n:longint;begin readln(a,b); c:=a mod b; while c<>0 do begin a:=b; b:=c; c:=a mod b; end; writeln(b);end.NUM.TWO

NUM.ONE——dijkstra单源最短路var a:array[1..100,1..100]of longint; i,j,k,l,m,n,max,maxp:longint; dist:array[1..100]of longint; v:array[1..1000]of boolean;begin readln(n); for i:=1 to n do

var a:array[1..1000]of longint; n,i,j:longint;procedure qp(l,r:longint);var m,i,j,t:longint;begin i:=l; j:=r; m:=(i+j)div 2; repeat while a[i]>a[m] do inc(i); while a[j]<a[m] do

NUM.ONE——排列组合var m,n,i,j,k,p,q:longint;function f(x:longint):longint;var t:longint; begin t:=1; for i:=1 to x do t:=t*i; exit(t); end;begin read(m,n); writeln(f(m)div(f(m-n)*f(n)

NUM.ONE——PRIMvar a:array[1..1000,1..1000]of longint; i,j,k,l,m,n,p,min,minp:longint; v:array[1..1000]of boolean; d:array[1..1000]of longint;begin readln(n); for i:=1 to n do for j:

NUM.ONE——高精乘var a,b,c:array[1..100]of longint; s,s1,s2:ansistring; i,j,k,l,m,n:longint;begin readln(s1); readln(s2); if length(s1)<length(s2) then begin s:=s1;s1:=s2;s2:=s;