标签:
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
///1085422276 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<queue> #include<cmath> #include<map> #include<bitset> #include<set> #include<vector> using namespace std ; typedef long long ll; #define mem(a) memset(a,0,sizeof(a)) #define meminf(a) memset(a,127,sizeof(a)); #define TS printf("111111\n"); #define FOR(i,a,b) for( int i=a;i<=b;i++) #define FORJ(i,a,b) for(int i=a;i>=b;i--) #define READ(a,b,c) scanf("%d%d%d",&a,&b,&c) #define mod 1000000007 #define inf 100000 inline ll read() { ll x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘)f=-1; ch=getchar(); } while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘; ch=getchar(); } return x*f; } //**************************************** struct ss { int id,x; bool operator < (const ss &A)const { return id < A.id; } }; #define maxn 100000+5 set<ss >s; set<ss >::iterator it; int main() { int oo=1; ll n,q,m,x[maxn],op[maxn],vis[maxn],ans[maxn],A[maxn]; int T=read(); while(T--) { scanf("%I64d%I64d",&n,&m); FOR(i,1,n) { scanf("%I64d%I64d",&op[i],&x[i]); } mem(vis); FOR(i,1,n) { if(op[i]==2)vis[x[i]]=1; } mem(ans); ans[0]=1; FOR(i,1,n) { ans[i]=ans[i-1]; if(op[i]==1&&!vis[i]) { ans[i]=(ans[i]*x[i])%m; } } //cout<<ans[10]<<endl; s.clear(); for(int i=n; i>=1; i--) { ll tmp=1; for(it=s.begin(); it!=s.end(); it++) { if((*it).id>i)break; tmp*=(*it).x; tmp%=m; } A[i]=(ans[i]*tmp)%m; if(op[i]==2) { ss g; g.id=x[i]; g.x=x[x[i]]; s.insert(g); } } printf("Case #%d:\n",oo++); for(int i=1; i<=n; i++) cout<<A[i]<<endl; } return 0; }
HDU 5475An easy problem 离线set/线段树
标签:
原文地址:http://www.cnblogs.com/zxhl/p/4847701.html