标签：cto max math chmod java ref 代码 mat style
整理代码。。。
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2304 Accepted Submission(s): 818
水题
代码:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<cstdlib> 7 #include<queue> 8 #include<stack> 9 #include<vector> 10 using namespace std; 11 typedef long long ll; 12 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 13 const int maxn=25; 14 const int mod=1e9+7; 15 const int inf=0x3f3f3f3f; 16 const double eps=acos(-1.0); 17 int a[maxn],b[maxn],c[maxn],d[maxn]; 18 int main(int argc,const char *argv[]){ 19 string str1,str2,str3,str4; 20 int len1,len2,len3,len4; 21 int up,n; 22 cin>>n; 23 while(n--){ 24 cin>>str1>>str2>>str3>>str4; 25 len1=str1.length();len2=str2.length(); 26 len3=str3.length();len4=str4.length(); 27 memset(a,0,sizeof(a));memset(b,0,sizeof(b)); 28 int i,j,k; 29 for(i=len1-1,k=0;i!=-1;--i){ 30 a[k]=str1[i]-‘0‘; 31 k++; 32 } 33 for(j=len2-1,k=0;j!=-1;--j){ 34 b[k]=str2[j]-‘0‘; 35 k++; 36 } 37 for(i=0,up=0;i<25;++i){ 38 a[i]=a[i]+b[i]+up; 39 up=a[i]/10; 40 a[i]%=10; 41 } 42 memset(b,0,sizeof(b));memset(c,0,sizeof(c)); 43 for(i=len3-1,k=0;i!=-1;--i){ 44 b[k]=str3[i]-‘0‘; 45 k++; 46 } 47 for(j=len4-1,k=0;j!=-1;--j){ 48 c[k]=str4[j]-‘0‘; 49 k++; 50 } 51 for(int i=0,up=0;i<maxn;++i){ 52 b[i]=b[i]+c[i]+up; 53 up=b[i]/10; 54 b[i]%=10; 55 } 56 for(i=0,up=0;i<maxn;++i){ 57 a[i]=a[i]+b[i]+up; 58 up=a[i]/10; 59 a[i]%=10; 60 } 61 for(i=maxn-1;i>=0;i--){ 62 if(a[i])break; 63 } 64 if(i==-1) 65 cout<<"0"; 66 else 67 for(;i>=0;i--) 68 cout<<a[i]; 69 cout<<endl; 70 } 71 return 0; 72 }
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1193 Accepted Submission(s): 628
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=1e5+10; 4 const int INF=0x3f3f3f3f; 5 int flag[N]; 6 int a[N]; 7 int main(){ 8 int t,n,minn; 9 int num1,num2,num; 10 while(~scanf("%d",&t)){ 11 while(t--){ 12 scanf("%d",&n); 13 memset(flag,0,sizeof(flag)); 14 for(int i=0;i<n;i++){ 15 scanf("%d",&a[i]); 16 flag[a[i]]=1; 17 } 18 num1=0,num2=0,num=0; 19 for(int i=a[0];i<=a[1];i++){ 20 if(flag[i]==0)num1++; 21 } 22 for(int i=a[n-2];i<=a[n-1];i++){ 23 if(flag[i]==0)num2++; 24 } 25 for(int i=a[0];i<=a[n-1];i++){ 26 if(flag[i]==0)num++; 27 } 28 //cout<<num1<<" "<<num2<<" "<<num<<endl; 29 minn=min(num1,num2); 30 printf("%d\n",num-minn); 31 } 32 } 33 return 0; 34 }
2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）HDU6225.Little Boxes-大数加法
标签：cto max math chmod java ref 代码 mat style
原文地址：https://www.cnblogs.com/ZERO-/p/9129905.html
