高精度加法

輸入兩個數，輸出他們的和（高精度）

輸入樣例

111111111111111111111111111111 222222222222222222222222222222

輸出樣例

333333333333333333333333333333

#include <bits/stdc++.h>
using namespace std;string a,b;
int aa[510],bb[510];
int c[510];int main()
{cin>>a>>b;int la = a.size();int lb = b.size();for(int i = la-1;i>=0;i--){aa[la-1-i] = a[i]-'0';}for(int i = lb-1;i>=0;i--){bb[lb-1-i] = b[i]-'0';}for(int i = 0;i<max(la,lb);i++){c[i] = aa[i]+bb[i];}for(int i = 0;i<max(la,lb);i++){if(c[i]>=10){c[i+1]++;}c[i] = c[i]%10;}for(int i = max(la,lb);i>=0;i--){if(i==max(la,lb)){if(c[i]!=0) cout<<c[i];}else cout<<c[i];}return 0;
}

高精度減法

輸入兩個數，輸出他們的差（高精度）

輸入樣例

333333333333333333333333333333 222222222222222222222222222222

輸出樣例

111111111111111111111111111111

#include <bits/stdc++.h>
using namespace std;string a,b;
int aa[10090],bb[10090];
int c[10090];int main()
{cin>>a>>b;int la = a.size();int lb = b.size();for(int i = la-1;i>=0;i--){aa[la-i] = a[i]-'0';}for(int i = lb-1;i>=0;i--){bb[lb-i] = b[i]-'0';}int jin = 0;bool fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff = false;if(la>lb){fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff = true;}else if(la==lb){for(int i = 0;i<la;i++){if(a[i]>b[i]){fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff = true;break;}if(a[i]<b[i]) break;}}if(fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff==true){for(int i = 1;i<=max(la,lb)+1;i++){int ccc = aa[i]-bb[i];c[i] = ccc%10;if(jin==1){c[i]--;jin = 0;if(c[i]<0){c[i] = (10+c[i])%10;jin = 1;}}if(c[i]<0){c[i] = (10+c[i])%10;jin = 1;}if(ccc<0) jin = 1;
//			cout<<i<<" "<<aa[i]<<" "<<bb[i]<<" "<<c[i]<<endl;}}else{for(int i = 1;i<=max(la,lb)+1;i++){int ccc = bb[i]-aa[i];c[i] = ccc%10;if(jin==1){c[i]--;jin = 0;if(c[i]<0){c[i] = (10+c[i])%10;jin = 1;}}if(c[i]<0){c[i] = (10+c[i])%10;jin = 1;}if(ccc>=10) jin = 1;
//		cout<<i<<" "<<aa[i]<<" "<<bb[i]<<" "<<c[i]<<endl;}}if(c[0]==0&&c[1]==0&&c[3]==0&&c[4]==0&&c[5]==0&&c[6]==0&&c[7]==0&&c[8]==0&&c[9]==0&&c[10]==0&&c[11]==0&&c[12]==0&&c[13]==0&&c[15]==0){cout<<0;return 0;}if(fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff==false) cout<<"-";for(int i = max(la,lb)+1;i>=1;i--){if(i==max(la,lb)+1){if(c[i]!=0) cout<<c[i];}else cout<<c[i];}return 0;
}

高精度乘法

輸入兩個數，輸出他們的積（高精度）

輸入樣例

11 11

輸出樣例

121

#include <bits/stdc++.h>
using namespace std;string a,b;
int aa[2010],bb[2010];
int c[4020];int main()
{cin>>a>>b;if(a=="0"||b=="0"){cout<<0;return 0;}int la = a.size();int lb = b.size();for(int i = la-1;i>=0;i--){aa[la-i] = a[i]-'0';}for(int i = lb-1;i>=0;i--){bb[lb-i] = b[i]-'0';}for(int i = 1;i<=lb;i++){for(int j = 1;j<=la;j++){c[j+i-1] += aa[j]*bb[i];}}
//	for(int i = 1;i<=la+lb;i++)
//	{
//		cout<<i<<" "<<c[i]<<endl;
//	}
//	cout<<endl<<endl<<endl;int jin = 0;for(int i = 1;i<=la+lb;i++){int ccccc = c[i]+jin;c[i] = ccccc%10;jin = ccccc/10;}for(int i = la+lb;i>=1;i--){if((i==la+lb&&c[i]!=0)||i!=la+lb){cout<<c[i];}}return 0;
}