题解 - Vijos

-1

天道无极 LV 5 @ 2015-09-16 20:14:34

var a,b:longint;
begin
readln(a,b);
writeln(a+b);
end.
PACAL算法！。

-1

zhaizhiyuan LV 4 @ 2015-09-12 15:20:32

#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
return 0;
}

-1

limuye LV 9 @ 2015-08-14 11:35:15

AC
#include <cstdio>
int main()
{
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", a + b);
return 0;
}

-1

pengyao1207 LV 7 @ 2015-07-31 19:24:36

记录信息
评测状态 Accepted
题目 P1000 A+B Problem
递交时间 2015-07-31 19:23:11
代码语言 C++
评测机 VijosEx
消耗时间 22 ms
消耗内存 532 KiB
评测时间 2015-07-31 19:23:15

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #3: Accepted, time = 1 ms, mem = 528 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 532 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #8: Accepted, time = 6 ms, mem = 524 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 524 KiB, score = 10

Accepted, time = 22 ms, mem = 532 KiB, score = 100

代码
#include<iostream>
#include<string>
//定义精度位数常量<255
using namespace std;
int strbj(string aa,string bb)
//字符串比较
{
int la=aa.length();
int lb=bb.length();
if(la>lb)return 1;
if(la<lb)return 0;
if(aa>bb)return 1;
return 0;
}
string gjdqh(string a,string b)
//高精度加法c++,by pengyao1207
{

int t=0;
int jw=0;
int la=a.length();
int lb=b.length();
string c="";
int ka=la-1;
int kb=lb-1;
int km=(la<lb?la:lb);
int ll=(la>lb)?(la-lb):(lb-la);
for(int i=1;i<=km;i++)
{

t=a[ka]+b[kb]-96+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
ka--;
kb--;
}
if(la>lb)
for(int i=1;i<=ll;i++)
{
t=a[ka]-48+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
ka--;
}
else
for(int i=1;i<=ll;i++)
{
t=b[kb]-48+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
kb--;
}
if(jw==1)c="1"+c;
return c;
}
string gjdqc(string a,string b)
//高精度减法c++,by pengyao1207
{

int la=a.length();
int lb=b.length();
string c="";
int ka=la-1;
int kb=lb-1;
int km=lb;
int ll=la-lb;
for(int i=1;i<=km;i++)
{
if(a[ka]<b[kb])
{
a[ka]=a[ka]+10;
a[ka-1]--;
}
c.insert(0,1,char(a[ka]-b[kb]+48));
ka--;
kb--;
}
for(int i=1;i<=ll;i++)
{
if(a[ka]<48)
{
c.insert(0,1,'9');
a[ka-1]--;
}
else c.insert(0,1,a[ka]);
ka--;
}
while((c[0]=='0')&&(c.length()!=1))c.erase(0,1);
return c;
}
string gjdqj(string a,string b)
//高精度乘法c++,by pengyao1207
{

string str="0";
string fz[10];
int lb=b.length();
int kb=lb-1;
fz[0]="0";
fz[1]=a;
fz[2]=gjdqh(fz[1],a);
fz[3]=gjdqh(fz[2],a);
fz[4]=gjdqh(fz[3],a);
fz[5]=gjdqh(fz[4],a);
fz[6]=gjdqh(fz[5],a);
fz[7]=gjdqh(fz[6],a);
fz[8]=gjdqh(fz[7],a);
fz[9]=gjdqh(fz[8],a);
int sl;
for(int i=0;i<=kb;i++)
{

string str1="0";
str1=fz[b[i]-48];
sl=kb-i;
//for(int k=1;k<=sl;k++)str1=str1+"0";
str1.insert(str1.length(),sl,'0');
str=gjdqh(str1,str);
}
while((str[0]=='0')&&(str.length()!=1))str.erase(0,1);
return str;
}
string gjdqs(string a,string b)
//高精度除法c++,by pengyao1207
{

int la=a.length();
int ka=la-1;
string str="";
string ys="0";
int zds;
string fz[10];
fz[0]="0";
fz[1]=b;
fz[2]=gjdqh(fz[1],b);
fz[3]=gjdqh(fz[2],b);
fz[4]=gjdqh(fz[3],b);
fz[5]=gjdqh(fz[4],b);
fz[6]=gjdqh(fz[5],b);
fz[7]=gjdqh(fz[6],b);
fz[8]=gjdqh(fz[7],b);
fz[9]=gjdqh(fz[8],b);
for(int i=0;i<=ka;i++)
{

if(ys=="0")ys=a[i];

else
{
ys.insert(ys.length(),1,'0');
ys=gjdqh(ys,a.substr(i,1));
}
zds=9;
for(int j=1;j<=9;j++)
{

if(strbj(fz[j],ys))
{
zds=j-1;
break;
}
}
str.insert(str.length(),1,char(zds+48));
ys=gjdqc(ys,fz[zds]);
}
while((str[0]=='0')&&(str.length()!=1))str.erase(0,1);
return str;
}
int main()
{
string a,b;
cin>>a>>b;
cout<<gjdqh(a,b);
}
记录信息
评测状态 Accepted
题目 P1000 A+B Problem
递交时间 2015-07-31 19:23:11
代码语言 C++
评测机 VijosEx
消耗时间 22 ms
消耗内存 532 KiB
评测时间 2015-07-31 19:23:15

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #3: Accepted, time = 1 ms, mem = 528 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 532 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #8: Accepted, time = 6 ms, mem = 524 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 524 KiB, score = 10

Accepted, time = 22 ms, mem = 532 KiB, score = 100

代码
#include<iostream>
#include<string>
//定义精度位数常量<255
using namespace std;
int strbj(string aa,string bb)
//字符串比较
{
int la=aa.length();
int lb=bb.length();
if(la>lb)return 1;
if(la<lb)return 0;
if(aa>bb)return 1;
return 0;
}
string gjdqh(string a,string b)
//高精度加法c++,by pengyao1207
{

int t=0;
int jw=0;
int la=a.length();
int lb=b.length();
string c="";
int ka=la-1;
int kb=lb-1;
int km=(la<lb?la:lb);
int ll=(la>lb)?(la-lb):(lb-la);
for(int i=1;i<=km;i++)
{

t=a[ka]+b[kb]-96+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
ka--;
kb--;
}
if(la>lb)
for(int i=1;i<=ll;i++)
{
t=a[ka]-48+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
ka--;
}
else
for(int i=1;i<=ll;i++)
{
t=b[kb]-48+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
kb--;
}
if(jw==1)c="1"+c;
return c;
}
string gjdqc(string a,string b)
//高精度减法c++,by pengyao1207
{

int la=a.length();
int lb=b.length();
string c="";
int ka=la-1;
int kb=lb-1;
int km=lb;
int ll=la-lb;
for(int i=1;i<=km;i++)
{
if(a[ka]<b[kb])
{
a[ka]=a[ka]+10;
a[ka-1]--;
}
c.insert(0,1,char(a[ka]-b[kb]+48));
ka--;
kb--;
}
for(int i=1;i<=ll;i++)
{
if(a[ka]<48)
{
c.insert(0,1,'9');
a[ka-1]--;
}
else c.insert(0,1,a[ka]);
ka--;
}
while((c[0]=='0')&&(c.length()!=1))c.erase(0,1);
return c;
}
string gjdqj(string a,string b)
//高精度乘法c++,by pengyao1207
{

string str="0";
string fz[10];
int lb=b.length();
int kb=lb-1;
fz[0]="0";
fz[1]=a;
fz[2]=gjdqh(fz[1],a);
fz[3]=gjdqh(fz[2],a);
fz[4]=gjdqh(fz[3],a);
fz[5]=gjdqh(fz[4],a);
fz[6]=gjdqh(fz[5],a);
fz[7]=gjdqh(fz[6],a);
fz[8]=gjdqh(fz[7],a);
fz[9]=gjdqh(fz[8],a);
int sl;
for(int i=0;i<=kb;i++)
{

string str1="0";
str1=fz[b[i]-48];
sl=kb-i;
//for(int k=1;k<=sl;k++)str1=str1+"0";
str1.insert(str1.length(),sl,'0');
str=gjdqh(str1,str);
}
while((str[0]=='0')&&(str.length()!=1))str.erase(0,1);
return str;
}
string gjdqs(string a,string b)
//高精度除法c++,by pengyao1207
{

int la=a.length();
int ka=la-1;
string str="";
string ys="0";
int zds;
string fz[10];
fz[0]="0";
fz[1]=b;
fz[2]=gjdqh(fz[1],b);
fz[3]=gjdqh(fz[2],b);
fz[4]=gjdqh(fz[3],b);
fz[5]=gjdqh(fz[4],b);
fz[6]=gjdqh(fz[5],b);
fz[7]=gjdqh(fz[6],b);
fz[8]=gjdqh(fz[7],b);
fz[9]=gjdqh(fz[8],b);
for(int i=0;i<=ka;i++)
{

if(ys=="0")ys=a[i];

else
{
ys.insert(ys.length(),1,'0');
ys=gjdqh(ys,a.substr(i,1));
}
zds=9;
for(int j=1;j<=9;j++)
{

if(strbj(fz[j],ys))
{
zds=j-1;
break;
}
}
str.insert(str.length(),1,char(zds+48));
ys=gjdqc(ys,fz[zds]);
}
while((str[0]=='0')&&(str.length()!=1))str.erase(0,1);
return str;
}
int main()
{
string a,b;
cin>>a>>b;
cout<<gjdqh(a,b);
}
记录信息
评测状态 Accepted
题目 P1000 A+B Problem
递交时间 2015-07-31 19:23:11
代码语言 C++
评测机 VijosEx
消耗时间 22 ms
消耗内存 532 KiB
评测时间 2015-07-31 19:23:15

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #3: Accepted, time = 1 ms, mem = 528 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 532 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #8: Accepted, time = 6 ms, mem = 524 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 524 KiB, score = 10

Accepted, time = 22 ms, mem = 532 KiB, score = 100

代码
#include<iostream>
#include<string>
//定义精度位数常量<255
using namespace std;
int strbj(string aa,string bb)
//字符串比较
{
int la=aa.length();
int lb=bb.length();
if(la>lb)return 1;
if(la<lb)return 0;
if(aa>bb)return 1;
return 0;
}
string gjdqh(string a,string b)
//高精度加法c++,by pengyao1207
{

int t=0;
int jw=0;
int la=a.length();
int lb=b.length();
string c="";
int ka=la-1;
int kb=lb-1;
int km=(la<lb?la:lb);
int ll=(la>lb)?(la-lb):(lb-la);
for(int i=1;i<=km;i++)
{

t=a[ka]+b[kb]-96+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
ka--;
kb--;
}
if(la>lb)
for(int i=1;i<=ll;i++)
{
t=a[ka]-48+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
ka--;
}
else
for(int i=1;i<=ll;i++)
{
t=b[kb]-48+jw;
c.insert(0,1,char((t%10)+48));
jw=((t>9)?1:0);
kb--;
}
if(jw==1)c="1"+c;
return c;
}
string gjdqc(string a,string b)
//高精度减法c++,by pengyao1207
{

int la=a.length();
int lb=b.length();
string c="";
int ka=la-1;
int kb=lb-1;
int km=lb;
int ll=la-lb;
for(int i=1;i<=km;i++)
{
if(a[ka]<b[kb])
{
a[ka]=a[ka]+10;
a[ka-1]--;
}
c.insert(0,1,char(a[ka]-b[kb]+48));
ka--;
kb--;
}
for(int i=1;i<=ll;i++)
{
if(a[ka]<48)
{
c.insert(0,1,'9');
a[ka-1]--;
}
else c.insert(0,1,a[ka]);
ka--;
}
while((c[0]=='0')&&(c.length()!=1))c.erase(0,1);
return c;
}
string gjdqj(string a,string b)
//高精度乘法c++,by pengyao1207
{

string str="0";
string fz[10];
int lb=b.length();
int kb=lb-1;
fz[0]="0";
fz[1]=a;
fz[2]=gjdqh(fz[1],a);
fz[3]=gjdqh(fz[2],a);
fz[4]=gjdqh(fz[3],a);
fz[5]=gjdqh(fz[4],a);
fz[6]=gjdqh(fz[5],a);
fz[7]=gjdqh(fz[6],a);
fz[8]=gjdqh(fz[7],a);
fz[9]=gjdqh(fz[8],a);
int sl;
for(int i=0;i<=kb;i++)
{

string str1="0";
str1=fz[b[i]-48];
sl=kb-i;
//for(int k=1;k<=sl;k++)str1=str1+"0";
str1.insert(str1.length(),sl,'0');
str=gjdqh(str1,str);
}
while((str[0]=='0')&&(str.length()!=1))str.erase(0,1);
return str;
}
string gjdqs(string a,string b)
//高精度除法c++,by pengyao1207
{

int la=a.length();
int ka=la-1;
string str="";
string ys="0";
int zds;
string fz[10];
fz[0]="0";
fz[1]=b;
fz[2]=gjdqh(fz[1],b);
fz[3]=gjdqh(fz[2],b);
fz[4]=gjdqh(fz[3],b);
fz[5]=gjdqh(fz[4],b);
fz[6]=gjdqh(fz[5],b);
fz[7]=gjdqh(fz[6],b);
fz[8]=gjdqh(fz[7],b);
fz[9]=gjdqh(fz[8],b);
for(int i=0;i<=ka;i++)
{

if(ys=="0")ys=a[i];

else
{
ys.insert(ys.length(),1,'0');
ys=gjdqh(ys,a.substr(i,1));
}
zds=9;
for(int j=1;j<=9;j++)
{

if(strbj(fz[j],ys))
{
zds=j-1;
break;
}
}
str.insert(str.length(),1,char(zds+48));
ys=gjdqc(ys,fz[zds]);
}
while((str[0]=='0')&&(str.length()!=1))str.erase(0,1);
return str;
}
int main()
{
string a,b;
cin>>a>>b;
cout<<gjdqh(a,b);
}

-1

geagerg LV 7 @ 2015-07-25 21:39:08

#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
return 0;
}

-1

geagerg LV 7 @ 2015-07-25 21:35:00

#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
return 0;
}

-1

Snake_Amadeus LV 4 @ 2015-07-22 16:55:02

#define POI <cstdio>

#define Poi 0

#define poooi main()

#define poi int

#define poipoi ,

#define poipoipoi ;

#define poipoipoipoi

#define poipoipoipoipoi scanf

#define poipoipoipoipoipoi printf

#define poipoipoipoipoipoipoi return

#define poipoipoipoipoipoipoipoi "%d%d"

#define poipoipoipoipoipoipoipoipoi "%d"

#define poipoipoipoipoipoipoipoipoipoi (

#define poipoipoipoipoipoipoipoipoipoipoi )

#define poipoipoipoipoipoipoipoipoipoipoipoi +

#define poipoipoipoipoipoipoipoipoipoipoipoipoi &

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoi a

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoipoi b

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi {

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi }

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi using

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi namespace

#define poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi std

#include POI

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoi

poi

poipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoi

poi

poooi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoi

poipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoi

poipoi

poipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoi

poipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoi

poipoipoi

poipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoi

poipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

poipoipoipoipoipoipoipoipoipoipoi

poipoipoi

poipoipoipoipoipoipoi

Poi

poipoipoi

poipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoipoi

-1

numlock LV 7 @ 2015-07-14 11:18:21

#include<cstdio>
int pre[2000],sum[2000],next[2000],mis[200];
int x,n,t,m,y,k,a,b;
bool pd[200];
bool aa,bb;
int ax,ay,az,bx,by,bz,cx,cy,cz;
int max(int a,int b,int c=-100)
{
if(a>=b && a>=c) return a;
if(b>=a && b>=c) return b;
return c;
}
void swap()
{
if(x>y) {int p=x;x=y;y=p;}
}
void build(int i,int l,int r)
{
if (l==r) {int p;scanf("%d",&p);sum[i]=pre[i]=mis[i]=next[i]=p;return;}
int mid=l+r>>1;
build(i<<1,l,mid);
build(i<<1|1,mid+1,r);
mis[i]=max(mis[i<<1],mis[i<<1|1],next[i<<1]+pre[i<<1|1]);
pre[i]=max(pre[i<<1],sum[i<<1]+pre[i<<1|1]);
next[i]=max(next[i<<1|1],next[i<<1]+sum[i<<1|1]);
sum[i]=sum[i<<1]+sum[i<<1|1];
}
void change(int i,int x,int z,int l,int r)
{
if(l==r){pre[i]=mis[i]=next[i]=sum[i]=z;return;}
int mid=l+r>>1;
if(x<=mid) change(i<<1,x,z,l,mid);
else change(i<<1|1,x,z,mid+1,r);
mis[i]=max(mis[i<<1],mis[i<<1|1],next[i<<1]+pre[i<<1|1]);
pre[i]=max(pre[i<<1],sum[i<<1]+pre[i<<1|1]);
next[i]=max(next[i<<1|1],next[i<<1]+sum[i<<1|1]);
sum[i]=sum[i<<1]+sum[i<<1|1];
}
void query(int i,int x,int y,int l,int r)
{
if(x<=l && r<=y)
{
if(!t) {ax=mis[i];bz=next[i];}
else {bx=mis[i];by=pre[i];bz=next[i];}
cx=max(ax,bx,az+by);
cz=max(bz,sum[i]+az);
cx=max(cx,cz);
ax=cx;
az=cz;
t++;
return;
}
int mid=l+r>>1;
if(x<=mid) query(i<<1,x,y,l,mid);
if(y>mid) query(i<<1|1,x,y,mid+1,r);
}
int main()
{
build(1,1,2);
query(1,1,2,1,2);
printf("%d",mis[1]);
}

J.C. @ 2015-07-14 14:36:48

cao

-1

sp7509 LV 4 @ 2015-06-14 15:31:03

#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
return 0;
}

-1

幕后黑手 LV 7 @ 2015-05-18 13:14:31

难度为9也，最难的题目了，我这个蒟蒻做不出来呢。。
大神们求教：
#include<iostream>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;
cout<<a+b;
return 0
}
为什么没有错？

-1

chengzi1998 LV 4 @ 2015-05-08 17:33:46

#include <stdio.h>
int main()
{
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", a + b);
return 0;
}

-1

dllp LV 9 @ 2014-12-19 16:36:15

#include<iostream>
using namespace std;

int main()
{
int a,b,c;
cin>>a>>b;
c=a+b;
cout<<c<<endl;
return 0;
}

-1

DavidDeng LV 8 @ 2014-12-06 13:28:10

###略有难度。。。啊哈哈哈哈哈###
##高精度##
测试数据 #0: Accepted, time = 0 ms, mem = 300 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 300 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 300 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 296 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 296 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 300 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 296 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 296 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 300 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 300 KiB, score = 10
Accepted, time = 0 ms, mem = 300 KiB, score = 100

#/*代码*/#
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;

int a[1002],b[1002],c[2005],L,L1,L2;
char str1[1002],str2[1002];
int main(){
cin>>str1>>str2;
L1=strlen(str1);
L2=strlen(str2);
L=(max(L1,L2));
for (int i=0;i<L1;i++) a[i]=str1[L1-i-1]-'0';
for (int i=0;i<L2;i++) b[i]=str2[L2-i-1]-'0';
for (int i=0;i<L;i++) c[i]=a[i]+b[i];
for (int i=0;i<L;i++)
if (c[i]>=10) c[i+1]++,c[i]-=10;
if (c[L]!=0)
for (int i=L;i>=0;i--) cout<<c[i];
else for (int i=L-1;i>=0;i--)
cout<<c[i];
return 0;
}

-1

zccz LV 5 @ 2014-11-04 21:15:57

###Plus A+B
#include <stdio.h>
int main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d",a+b);
return 0;
}

-1

XYSZ2013WSL LV 7 @ 2014-11-03 18:50:28

program Plus;
var a,b:longint;
begin
readln(a,b);
writeln(a+b);
end.

-1

乖孩子 LV 3 @ 2014-10-30 18:21:36

var
a,b:integer;
begin
rea(a,b);
writeln(a+b);
end.

-1

Lostsong (LZW007) LV 7 @ 2014-10-29 16:28:12

123 500

-1

1065484105 LV 6 @ 2014-10-29 13:11:47

var x,y,z:integer;
begin
readln(x,y);
z:=x+y;
write(z);
end.

-1

Whitehead LV 3 @ 2014-10-28 17:42:30

var a,b,c:longint;
begin
read(a,b);
c:=a+b;
writeln(c);
end.

-1

芜湖27中小袁 LV 4 @ 2014-10-27 20:48:43

#include<iostream>
using namespace std;
int main(){
int a,b;
cin>>a>>b;
cout<<a+b;
return 0;
}

题解

1309 条题解

A+B Problem

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题解

1309 条题解

A+B Problem

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