1.高精度运算_加法
AYYZOJ p1443
COGS p37
1 type 2 arr=array[1..200]of integer; 3 var 4 a,b:arr;i,la,lb:integer; n:string; 5 procedure add(a,b:arr;la,lb:integer); 6 var i,x,lc:integer; c:arr; 7 begin 8 i:=1; x:=0; 9 while (i<=la) or(i<=lb) do 10 begin x:=a[i]+b[i]+x; 11 c[i]:=x mod 10; 12 x:=x div 10 ; 13 i:=i+1; 14 end; 15 if x>0 then begin lc:=i; c[lc]:=x; end else lc:=i-1; 16 for i:=lc downto 1 do write(c[i]); 17 writeln; 18 end; 19 begin 20 readln(n); la:=length(n); 21 for i:=1 to la do a[i]:=ord(n[la-i+1])-ord('0'); 22 readln(n); lb:=length(n); 23 for i:=1 to lb do b[i]:=ord(n[lb-i+1])-ord('0'); 24 add(a,b,la,lb); 25 end.
2.高精度运算_减法
AYYZOJ p1444
COGS p38
1 type 2 arr=array[1..200]of integer; 3 var 4 a,b:arr;i,la,lb:integer; n,n1,n2:string; 5 procedure sub(a,b:arr;la,lb:integer); 6 var i,x,lc:integer; c:arr; 7 begin 8 i:=1; 9 x:=0; 10 fillchar(c,sizeof(c),0); 11 while (i<=la) or(i<=lb) do 12 begin 13 x := a[i] - b[i] + 10 + x; {不考虑大小问题，先往高位借10} 14 c[i] := x mod 10 ; {保存第i 位的值} 15 x := x div 10 - 1; {将高位借掉的1减去} 16 i := i + 1 17 end; 18 lc:=i; 19 while (c[lc]=0) and (lc>1) do dec(lc); {最高位的0 不输出} 20 for i:=lc downto 1 do write(c[i]); 21 writeln 22 end; 23 begin 24 readln(n1); 25 readln(n2); {处理被减数和减数} 26 la:=length(n1); lb:=length(n2); 27 if (la<lb) or (la=lb) and (n1<n2) then 28 begin 29 n:=n1;n1:=n2;n2:=n; 30 write('-') {n1<n2,结果为负数} 31 end; 32 la:=length(n1); lb:=length(n2); 33 for i:=1 to la do a[la-i+1]:=ord(n1[i])-ord('0'); 34 for i:=1 to lb do b[lb-i+1]:=ord(n2[i])-ord('0'); 35 sub(a,b,la,lb); 36 end.
3.高精度运算-数列求和
AYYZOJ p1448
(⊙v⊙)嗯 首先你需要知道这个：
好吧，其实你只需要得到：
野生的新鲜公式：
| 描述 Description | ||
| 费波那契数列的前两项分别为1,1。以后每项为前两项之和。 输入n,求费波那契数列前n项的和(1<=n<=5000)。 输入：仅一个数，n 输出：费波那契数列前n项之和。 Sample Input 3 Sample Output 4 ------------------------------- 对于样例的解释 费波那契数列前三项是1,1,2，和为4。 输入文件：fbnq.in 输出文件：fbnq.out |
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| 时间限制 Time Limitation | ||
| 各个测试点1s | ||
1 program fbnq; 2 type 3 arr=array[1..7010] of integer; 4 var 5 a,b,c:arr; 6 l,i,n:integer; 7 procedure plus(a1,b1:arr;var x:arr); 8 var i,k:integer; 9 begin 10 k:=0; 11 for i:=1 to l do 12 begin 13 a1[i]:=a1[i]+b1[i]+k; 14 k:=a1[i] div 10; 15 a1[i]:=a1[i] mod 10; 16 end; 17 if k>0 then begin inc(l);a1[l]:=k;end; 18 x:=a1; 19 end; 20 begin 21 readln(n); 22 fillchar(a,sizeof(a),0); 23 fillchar(b,sizeof(b),0); 24 fillchar(c,sizeof(c),0); 25 if n=1 then begin writeln(1);halt;end; 26 if n=2 then begin writeln(2);halt;end; 27 a[1]:=1; 28 b[1]:=1; 29 l:=1; 30 for i:=3 to n+2 do 31 begin 32 plus(a,b,c); 33 a:=b; 34 b:=c; 35 end; 36 if c[1]-1<0 then 37 begin 38 c[1]:=9; 39 dec(c[2]); 40 for i:=2 to l do 41 begin 42 if (c[i]<0) then begin c[i+1]:=c[i+1]-1;c[i]:=c[i]+10; end 43 else break; 44 end; 45 end 46 else dec(c[1]); 47 while c[l]=0 do dec(l); 48 for i:=l downto 1 do write(c[i]);writeln; 49 end.
1 program fbnq; 2 type 3 arr=array[1..7010] of integer; 4 var 5 a,b,c:arr; 6 l,i,n:integer; 7 function plus(a,b:arr):arr; 8 var i,c:integer; 9 begin 10 c:=0; 11 for i:=1 to l do 12 begin 13 a[i]:=a[i]+b[i]+c; 14 c:=a[i] div 10; 15 a[i]:=a[i] mod 10; 16 end; 17 if c>0 then begin inc(l);a[l]:=c;end; 18 plus:=a; 19 end; 20 begin 21 readln(n); 22 fillchar(a,sizeof(a),0); 23 fillchar(b,sizeof(b),0); 24 fillchar(c,sizeof(c),0); 25 if n=1 then begin writeln(1);halt;end; 26 if n=2 then begin writeln(2);halt;end; 27 a[1]:=1; 28 b[1]:=1; 29 l:=1; 30 for i:=3 to n+2 do 31 begin 32 c:=plus(a,b); 33 a:=b; 34 b:=c; 35 end; 36 if c[1]-1<0 then 37 begin 38 c[1]:=9; 39 dec(c[2]); 40 for i:=2 to l do 41 begin 42 if (c[i]<0) then begin c[i+1]:=c[i+1]-1;c[i]:=c[i]+10; end 43 else break; 44 end; 45 end 46 else dec(c[1]); 47 while c[l]=0 do dec(l); 48 for i:=l downto 1 do write(c[i]);writeln; 49 end.
4.NOIP1999_PT2_回文数
AYYZOJ p1008
COGS p40
1 program huiwen; 2 type arr=array[1..50] of byte; 3 var 4 a:arr; 5 n,step,l,i,j:integer; 6 m:string[20]; 7 procedure add(var a:arr); 8 var c,i:integer; 9 b:arr; 10 begin 11 b:=a; 12 c:=0; 13 for i:=1 to l do 14 begin 15 a[i]:=a[i]+b[l+1-i]+c; 16 c:=a[i] div n; 17 a[i]:=a[i] mod n; 18 end; 19 if c<>0 then begin inc(l);a[l]:=c; end; 20 end; 21 procedure try(var a:arr); 22 var i:integer; 23 begin 24 for i:= 1 to l div 2 do 25 if a[i]<>a[l+1-i] then exit; 26 writeln('STEP=',step); 27 halt; 28 end; 29 begin 30 readln(n); 31 readln(m); 32 l:=length(m); 33 for i:=1 to l do 34 if m[i] in ['0'..'9'] then a[l-i+1]:=ord(m[i])-ord('0') 35 else case m[i] of 36 'A','a':a[l-i+1]:=10; 37 'B','b':a[l-i+1]:=11; 38 'C','c':a[l-i+1]:=12; 39 'D','d':a[l-i+1]:=13; 40 'E','e':a[l-i+1]:=14; 41 'F','f':a[l-i+1]:=15; 42 end; 43 step:=0; 44 try(a); 45 while step<=30 do 46 begin 47 add(a); 48 inc(step); 49 try(a); 50 end; 51 writeln('Impossible!'); 52 end.