u013570474的专栏

#include #include const int N=1000000;int hash[2*N+1],a,b,c,d,ans;int main(){    int i,j;    while (scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF){        if ((a>0 && b>0 && c>0 && d

#include #include #include #include using namespace std;#define MAX 16000005char s[MAX];int ascii[256];int hash[MAX];int main(){    //freopen("acm.txt","r",stdin)

#include #include #include using namespace std;#define mod 1313struct node{    int d;    struct node*next;};node *pt[mod + 10];int a,b,c,d,e;int x,y,z,n,m;

数字去重和排序IITime Limit: 4000 MSMemory Limit: 65536 KTotal Submit: 555(131 users)Total Accepted: 234(116 users)Rating: Special Judge: NoDescription

#include #include #include #include #include #define mod 1313using namespace std;struct node{    int len[10];};bool cmp(node &a,node &b){ int start[6],i,j,num

Consider equations having the following form: a*x1^2+b*x2^2+c*x3^2+d*x4^2=0 a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0. It is consider a solution a system

#include #include #include #include #include #include using namespace std;int lev[10000];map ma;int main(){    int n;    while(scanf("%d",&n)==1)    {        ma.clear()

#include #include #include #include #include #include #include using namespace std;int ans;struct node{ int data1, data2; int next;};node hash1[1000005];int index1[100007];int a[1005],