【zkw线段树】ural1855

比赛时脑残了，想起去维护前缀和，结果被标记维护恶心到了，其实只要维护每一段就行了，比赛时一条路走到黑，没想起另开一条路

维护每一段就只要维护sigma(ai),sigma(i*ai),sigma(i*i*ai)即可.被long long坑了很久.zkw线段树依然给力地闯进了第一版.

最近几次试组队发现码力这东西确实相当重要，毕竟机时只有那么多，如何稳妥快速实现是很重要的。同时,想算法还要更加灵活，有时候想算法就是会脑残一下...

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;long long s[2000000][3],delt[2000000][3],t[2000000],ans[3];int n,m,m1,high;char ch[20000];void origin(){for (m1=1,high=0;m1<=n+2;m1<<=1,high++) ;for (int i=1;i<=n;i++) delt[i+m1][0]=1,delt[i+m1][1]=(i),delt[i+m1][2]=((long long)i)*(i);for (int i=m1-1;i>=1;i--) {int l=i<<1,r=l+1;delt[i][0]=delt[l][0]+delt[r][0];delt[i][1]=delt[l][1]+delt[r][1];delt[i][2]=delt[l][2]+delt[r][2];}}void change(int x,long long d){t[x]+=d;s[x][0]+=delt[x][0]*d;s[x][1]+=delt[x][1]*d;s[x][2]+=delt[x][2]*d;}void pushdown(int x){for (int i=high-1;i>=0;i--) {int ne=x>>i;if (t[ne]) {change(ne<<1,t[ne]);change((ne<<1)+1,t[ne]);t[ne]=0;}}}void updata(int x){for (;x;x>>=1) {s[x][0]=s[x<<1][0]+s[(x<<1)+1][0];s[x][1]=s[x<<1][1]+s[(x<<1)+1][1];s[x][2]=s[x<<1][2]+s[(x<<1)+1][2];}}void modify(int l,int r,long long d){if (l>r) return ;l+=m1-1,r+=m1+1;int ll=l>>1,rr=r>>1;pushdown(ll),pushdown(rr);for (;!((l^r)==1);l>>=1,r>>=1) {if ((l&1)==0) change(l+1,d);if ((r&1)==1) change(r-1,d);}updata(ll),updata(rr);}void ask(int l,int r,long long ans[3]){ans[0]=ans[1]=ans[2]=0;if (l>r) return ;l+=m1-1,r+=m1+1;int ll=l>>1,rr=r>>1;pushdown(ll),pushdown(rr);for (;!((l^r)==1);l>>=1,r>>=1) {if ((l&1)==0) {ans[0]+=s[l+1][0];ans[1]+=s[l+1][1];ans[2]+=s[l+1][2];}if ((r&1)==1) {ans[0]+=s[r-1][0];ans[1]+=s[r-1][1];ans[2]+=s[r-1][2];}}}int main(){freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);scanf("%d%d\n",&n,&m);origin();for (int i=1;i<=m;i++) {scanf("%s",ch+1);if ('c'==ch[1]) {int l,r,d;scanf("%d%d%d\n",&l,&r,&d);modify(l+1,r,d);/*printf("%lld\n",(r+l+1)*ans[1]-ans[2]-(l+l*r)*ans[0]);printf("%lld %lld %lld\n\n",ans[0],ans[1],ans[2]);*/}else {int l,r;scanf("%d%d\n",&l,&r);ask(l+1,r,ans);long long sum=((long long)r-l+1)*((long long)r-l)/2;//printf("%lld\n",(r+l+1)*ans[1]-ans[2]-(l+l*r)*ans[0]);//printf("%lld %lld %lld\n",ans[0],ans[1],ans[2]);if (sum) printf("%.10lf\n",((double)((long long)r+l+1)*ans[1]-ans[2]-((long long)l+((long long)l)*r)*ans[0])/sum);else printf("%.10lf\n",0.0);}}/*ask(2,2,ans);printf("%lld %lld %lld\n",ans[0],ans[1],ans[2]);*/return 0;}