主要是考虑分类讨论,代码比较繁琐
一直想减少情况还是到不了0ms
#pragma GCC optimize(2)
#include
int main()
{
int T;scanf("%d",&T);
while(T--){
int n,m;scanf("%d%d",&n,&m);
int num[10];
for(register int i=1;i<=m;i++) scanf("%d",num+i);
if(m==1){
if(num[1]==1) printf("1\n");
else printf("-1\n");
continue;
}
if(m==2){
if(n==2) printf("-1\n");
else if((num[1]==1&&num[2]-num[1]==1)) printf("1\n");
else if((num[2]==n&&num[2]-num[1]==1)) printf("%lld\n",n);
else printf("-1\n");
continue;
}
if(m==3){
if(num[2]-num[1]==1&&num[3]-num[2]==1) printf("%lld\n",num[2]);
else printf("-1\n");
continue;
}
}
return 0;
}
水题..
#pragma GCC optimize(2)
#include
int a[26];
int main()
{
read(n);
for(int i=0;i<26;i++) scanf("%d",a+i);
long long ans = 0;
for(int i=0;i<26;i++) ans += i/n;
printf("%lld\n",ans);
return 0;
}
这个题说一下吧。
建立一个虚根(!!这个想法可能会降低树的高度!!!)
把所有的已知解全部连接到该跟上
每一步的答案即为该跟的子树大小
具体细节看一下代码吧:
#include
#pragma GCC optimize(3)
using namespace std;
typedef long long ll;
const int maxn = 4e5+6;
templatevoid read(T &x)
{
x=0;int f=0;char ch=getchar();
while(ch<'0'||ch>'9') {f|=(ch=='-');ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
x=f?-x:x;return;
}
int pre[maxn],sz[maxn];
int Find(int x){
return pre[x]==x?x:pre[x]=Find(pre[x]);
}
int main(){
int ans = 0;
int n,m;read(n);read(m);
for(register int i=1;i<=n;i++) pre[i] = i,sz[i] = 1;
pre[n+1] = n+1;
for(register int i=1;i<=m;i++){
int op,x,y;
read(op);read(x);
if(op==1){
int dx = Find(x);
if(dx!=n+1) {
pre[dx] = n+1;
ans+=sz[dx];
}
}
else{
read(y);
int dx = Find(x),dy = Find(y);
if(dx != dy){
if(dx == n+1){
ans+=sz[dy];
pre[dy] = n+1;
}
else if(dy == n+1){
ans+=sz[dx];
pre[dx] = n+1;
}
else if(dx!=dy) pre[dx] = dy,sz[dy]+=sz[dx];
}
}
printf("%d\n",ans);
}
}