分块 （区间修改，区间查询
比较常规的操作，用树状数组和线段树等都行。复杂度就不分析了
题目链接https://loj.ac/problem/6279
1 #include<bits/stdc++.h> 2 3 4 using namespace std; 5 typedef long long ll; 6 const int maxn = 1e5+10; 7 const int inf = 0x3f3f3f3f; 8 int dir[10][10] = {{1,0},{-1,0},{0,1},{0,-1}}; 9 ll sum[maxn],a[maxn],lz[maxn]; 10 int opt,l,r,c,n,t,L[maxn],R[maxn],pos[maxn]; 11 12 inline int read() 13 { 14 char ch = getchar(); ll k = 0, f = 1; 15 while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} 16 while(ch >= '0' && ch <= '9') {k = (k<<1) + (k<<3) + ch - '0'; ch = getchar();} 17 return k * f; 18 } 19 20 inline void add(int l,int r,int c)//添加 21 { 22 if(pos[l] == pos[r]) 23 { 24 for(int i = l; i <= r; ++i) a[i] += c; 25 sum[pos[l]] += (r - l + 1) * c; 26 } 27 28 else 29 { 30 for(int i = l; i <= R[pos[l]]; ++i) a[i] += c; 31 sum[pos[l]] += (R[pos[l]] - l + 1) * c; 32 for(int i = L[pos[r]]; i <= r; ++i) a[i] += c; 33 sum[pos[r]] += (r - L[pos[r]] + 1) * c; 34 for(int i = pos[l] + 1; i <= pos[r] - 1; ++i) sum[i] += (R[i] - L[i] +1 ) * c, lz[i] += c; 35 } 36 37 } 38 39 inline ll findy(int l,int r,int c)//查询 40 { 41 ll ans = 0; 42 if(pos[l] == pos[r]) 43 { 44 for(int i = l; i <= r; ++i) ans += (a[i] + lz[pos[l]]), ans %= (c + 1); 45 } 46 47 else 48 { 49 for(int i = l; i <= R[pos[l]]; ++i) ans += (a[i] + lz[pos[l]]), ans %= (c + 1); 50 for(int i = L[pos[r]]; i <= r; ++i) ans += (a[i] + lz[pos[r]]), ans %= (c + 1); 51 for(int i = pos[l] + 1; i <= pos[r] - 1; ++i) 52 { 53 ans += sum[i], ans %= (c + 1); 54 } 55 56 } 57 return ans %= (c + 1); 58 } 59 60 int main() 61 { 62 n=read(); 63 for(int i = 1; i <= n; ++i) cin>>a[i]; 64 t = sqrt(n); 65 for(int i = 1; i <= t; ++i) 66 { 67 L[i] = (i-1) * t + 1; 68 R[i] = i * t; 69 } 70 71 if(R[t] != n) t++, L[t] = R[t - 1] + 1 ,R[t] = n; 72 for(int i = 1; i <= t; ++i) 73 { 74 for(int j = L[i]; j <= R[i]; ++j) 75 { 76 pos[j] = i; sum[i] += a[j]; 77 } 78 79 } 80 81 for(int i = 1; i <= n; ++i) 82 { 83 opt = read(); l = read(); r = read(); c=read(); 84 if(opt == 0) add(l, r, c); 85 else cout<<findy(l, r, c)<<endl; 86 87 } 88 return 0; 89 }
