LOJ #6281. 数列分块入门 5
区间开方，求和。
分析性质可发现sqrt（0）= 0，sqrt(1) = 1，所以记录每块是否已不能改变。
区分l与r啊。。错了这么个关键字母可不爆零了么。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 using namespace std; 7 typedef long long ll; 8 9 ll bl[50050],fns[250],v[50050],sum[250]; 10 ll n,blo,c,l,r,opt; 11 12 void check(ll x){ 13 for(register int i = (x-1)*blo+1;i <= min(n,x*blo);i++) 14 if(v[i]^0&&v[i]^1)return; 15 fns[x] = 1; 16 } 17 18 void change(ll l,ll r){ 19 ll x,flag; 20 if(bl[l] == bl[r]){ 21 flag = true; 22 for(register int i = l;i <= r;i++)if(v[i] > 1){ 23 x = sqrt(v[i]); 24 sum[bl[l]] -= v[i]-x; 25 v[i] = x; 26 if(x > 1)flag = false; 27 } if(flag)check(bl[l]); return; 28 } 29 if(!fns[bl[l]]){ 30 flag = true; 31 for(register int i = l;i <= bl[l]*blo;i++)if(v[i] > 1){ 32 x = sqrt(v[i]); 33 sum[bl[l]] -= v[i]-x; 34 v[i] = x; 35 if(x > 1)flag = false; 36 } if(flag)check(bl[l]); 37 } 38 if(!fns[bl[r]]){ 39 flag = true; 40 for(register int i = (bl[r]-1)*blo+1;i <= r;i++)if(v[i] > 1){ 41 x = sqrt(v[i]); 42 sum[bl[r]] -= v[i]-x; 43 v[i] = x; 44 if(x > 1)flag = false; 45 } if(flag)check(bl[r]); 46 } 47 for(register int i = bl[l]+1;i < bl[r];i++)if(!fns[i]){ 48 flag = true; 49 for(register int j = (i-1)*blo+1;j <= min(n,i*blo);j++)if(v[j] > 1){ 50 x = sqrt(v[j]); 51 sum[i] -= v[j]-x; 52 v[j] = x; 53 if(x > 1)flag = false; 54 } if(flag)check(i); 55 } 56 } 57 58 ll query(ll l,ll r){ 59 ll ans = 0; 60 if(bl[l] == bl[r]){ 61 for(register int i = l;i <= r;i++)ans += v[i]; 62 return ans; 63 } 64 for(register int i = l;i <= bl[l]*blo;i++)ans += v[i]; 65 for(register int i = (bl[r]-1)*blo+1;i <= r;i++)ans += v[i]; 66 for(register int i = bl[l]+1;i < bl[r];i++)ans += sum[i]; 67 return ans; 68 } 69 70 int main(){ 71 scanf("%lld",&n); blo = sqrt(n); 72 for(register int i = 1;i <= n;i++){ 73 scanf("%lld",&v[i]); 74 bl[i] = (i-1)/blo+1; 75 sum[bl[i]] += v[i]; 76 } 77 for(register int i = 1;i <= bl[n];i++)check(i); 78 for(register int i = 1;i <= n;i++){ 79 scanf("%lld%lld%lld%lld",&opt,&l,&r,&c); 80 switch(opt){ 81 case 0:change(l,r);break; 82 case 1:printf("%lld\n",query(l,r));break; 83 } 84 } 85 return 0; 86 }
