BZOJ1754: [Usaco2005 qua]Bull Math
1754: [Usaco2005 qua]Bull Math
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 374 Solved: 227
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Description
Bulls are so much better at math than the cows. They can multiply
huge integers together and get perfectly precise answers ... or so
they say. Farmer John wonders if their answers are correct. Help
him check the bulls' answers. Read in two positive integers (no
more than 40 digits each) and compute their product. Output it as
a normal number (with no extra leading zeros).
FJ asks that you do this yourself; don't use a special library
function for the multiplication.
输入两个数,输出其乘积
Input
* Lines 1..2: Each line contains a single decimal number.
Output
* Line 1: The exact product of the two input lines
Sample Input
11111111111111
1111111111
1111111111
Sample Output
12345679011110987654321
HINT
Source
题解:
呵呵,金组出这种题,防爆零呢吧。。。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 3 #include<cstdlib> 4 5 #include<cmath> 6 7 #include<cstring> 8 9 #include<algorithm> 10 11 #include<iostream> 12 13 #include<vector> 14 15 #include<map> 16 17 #include<set> 18 19 #include<queue> 20 21 #include<string> 22 23 #define inf 1000000000 24 25 #define maxn 10000 26 27 #define maxm 500+100 28 29 #define eps 1e-10 30 31 #define ll long long 32 33 #define pa pair<int,int> 34 35 #define for0(i,n) for(int i=0;i<=(n);i++) 36 37 #define for1(i,n) for(int i=1;i<=(n);i++) 38 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 40 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 42 43 #define mod 1000000007 44 45 using namespace std; 46 47 inline int read() 48 49 { 50 51 int x=0,f=1;char ch=getchar(); 52 53 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 54 55 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 56 57 return x*f; 58 59 } 60 int n,m,a[maxn],b[maxn],c[maxn]; 61 char ch; 62 63 int main() 64 65 { 66 67 freopen("input.txt","r",stdin); 68 69 freopen("output.txt","w",stdout); 70 71 ch=getchar(); 72 while(ch<='9'&&ch>='0')a[++n]=ch-'0',ch=getchar(); 73 for1(i,n>>1)swap(a[i],a[n+1-i]); 74 ch=getchar(); 75 while(ch<='9'&&ch>='0')b[++m]=ch-'0',ch=getchar(); 76 for1(i,m>>1)swap(b[i],b[m+1-i]); 77 for1(i,n) 78 for1(j,m) 79 c[i+j-1]+=a[i]*b[j]; 80 for1(i,n+m)c[i+1]+=c[i]/10,c[i]%=10; 81 int t=n+m; 82 while(!c[t])t--; 83 for3(i,t,1)printf("%d",c[i]); 84 printf("\n"); 85 86 return 0; 87 88 }