input
3
1,0 2,1
4,2,0 1,2,0
1,0 10,6,4,2,1
output
case #0:
1,0,1
case #1:
1,1,1,0
case #2:
1,0,0,0,0,1
answer
case #0:
1,0,1
case #1:
1,1,1,0
case #2：
1,0,0,0,0,1
checker’s comment
wrong answer: line 5 differ - expected: ‘case #2：’, found: ‘case #2:’
???????????????????????????????????????????????????????????????????????
题目可能有点误导，感觉位权并没有用上
using namespace std;
typedef vector BIG;
int issu[1000];
vector su;
int T;
BIG input(string aa)
{
BIG ret,tmp;
stringstream ss(aa);
int xx;
ss>>xx;
tmp.push_back(xx);
while(ss.get()==’,’)
{
ss>>xx;
tmp.push_back(xx);
}
for(int i=tmp.size()-1;i>=0;i–) ret.push_back(tmp[i]);
return ret;
}
BIG operator+(const BIG& aa,const BIG& bb)
{
BIG ret;
for(int i=0;i<min(aa.size(),bb.size());i++) ret.push_back(aa[i]+bb[i]);
if(aa.size()<bb.size())
{
for(int i=aa.size();i<bb.size();i++) ret.push_back(bb[i]);
}
else
{
for(int i=bb.size();i=su[ret.size()-1])
{
int old=ret.size()-1;
ret.push_back(ret[old]/su[old]);
ret[old]%=su[old];
}
return ret;
}
void print(const BIG& aa)
{
int ff=1;
for(int i=aa.size()-1;i>=0;i–)
if(ff) {ff=0;cout<<aa[i];}
else cout<<’,’<<aa[i];
cout<<endl;
return;
}
int main()
{
for(int i=2;i<=900;i++) issu[i]=1;
for(int i=2;i<=900;i++)
if(issu[i])
for(int j=i*2;j<=900;j+=i)
issu[j]=0;
for(int i=2;i<=900;i++) if(issu[i]) su.push_back(i);
cin>>T;
for(int step=0;step>s1>>s2;
BIG a1=input(s1),a2=input(s2);
BIG a3=a1+a2;
printf(“case #%d:\n”,step);
print(a3);
}
return 0;
}
就是高精度加法变形一下即可。
再套一个素数筛法。
希望对以后的人有用！
注意：
1. 每一位是满几进一
2. 读取数字要倒着读
3. 质数表不要写错
附C代码：
#include <stdio.h>
#include <string.h>
int main()
{
int T, I, weight[25]={2,3,5,7,11, 13,17,19,23,29, 31,37,41,43,47, 53,59,61,67,71, 73,79,83,89,97};//满weight[i]进一
scanf("%d",&T);
for(I=0; I<T; I++)
{
int n1[26]={0}, n2[26]={0}, i, j, l1, l2, cnt;
char s1[100], s2[100];
scanf("%s %s",s1,s2);
l1=strlen(s1);
l2=strlen(s2);
for(i=l1-1, cnt=0, j=1; i+1; i--)//读第一个数
{
if(s1[i]==',') {j=1; cnt++;}
else
{
n1[cnt]+=(s1[i]-'0')*j;
j*=10;
}
}
for(i=l2-1, cnt=0, j=1; i+1; i--)//读第二个数
{
if(s2[i]==',') {j=1; cnt++;}
else
{
n2[cnt]+=(s2[i]-'0')*j;
j*=10;
}
}
for(i=0; i<25; i++)//模拟运算
{
n1[i]+=n2[i];
if(n1[i]>=weight[i])
{
n1[i+1]+=n1[i]/weight[i];
n1[i]%=weight[i];
}
}
printf("case #%d:\n",I);
for(i=25; i; i--) if(n1[i]) break;
for(; i; i--) printf("%d,",n1[i]);
printf("%d\n",n1[i]);
}
return 0;
}
from math import sqrt
def isPrime(n):
for i in range(2,int(sqrt(n))+1):
if n % i == 0:
return False
return True
l = [] #质数表
for x in range(2,103):
if isPrime(x):
l.append(x)
times = eval(input())
for j in range(times):
s = input()
a,b = s.split()
a = a.split(“,”)
b = b.split(“,”)
bitsa = len(a)
bitsb = len(b)
a1 = []
b1 = []
c1 = []
for i in range(26):
c1.append(0)