Topic links: https://vjudge.net/contest/332656#problem/G
Meaning of the questions:
Give you three operations, three output mode:
1. Let l ~ r interval value plus all C;
2. Let l ~ r C interval by all of the values;
3. Let l ~ r value becomes a range of all C;
4. Let l ~ r section to find all the values p (1 ~ 3) power, and then summed output.
Ideas:
This question is asked three pit: set, add, mul. So lazy numerals have three, if the processing method of simultaneously occurring three markers - When set the update operation, and put the add tag mul Mark all canceled; mul when updating operation, if the current node add marker is present, put add marking to: add * mul. In this case we can first perform in PushDown () operation set, then mul, and finally add.
sum1 = a1 + a2 + a3;
sum2 = a1 ^ 2 + a2 ^ 2 + a3 ^ 2;
sum3 = a1 ^ 3 + a2 ^ 3 + a3 ^ 3;
If each element c add it?
sum1 = a1 + a2 + a3 + 3*c;
sum2 = (a1 + c)^2 + (a2 + c) ^2 + (a3 + c)^2;
sum3 = (a1 + c)^3 + (a2 + c) ^3 + (a3 + c)^3;
Formulas decomposition does not look like
sum2 = a1^2 + a2^2 + a3^2 + 2 * c * (a + b + c) + 3 * c^2;
sum1 sum2 + 2 * c on the sum2 = a * a state in a state + 3 * c ^ 2;
sum3 = .....
C multiply each element, this is very simple, we know everything, every element becomes c is also very good operation.
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <cmath> 5 #include <string> 6 #include <string.h> 7 #include <algorithm> 8 using namespace std; 9 #define LL __int64 10 #define eps 1e-8 11 #define INF INT_MAX 12 #define lson l , m , rt << 1 13 #define rson m + 1 , r , rt << 1 | 1 14 const int MOD = 10007; 15 const int maxn = 100000 + 5; 16 const int N = 12; 17 LL add[maxn << 2] , set[maxn << 2] , mul[maxn << 2]; 18 LL sum1[maxn << 2] , sum2[maxn << 2] , sum3[maxn << 2]; 19 void PushUp(int rt) 20 { 21 sum1[rt] = (sum1[rt << 1] + sum1[rt << 1 | 1]) % MOD; 22 sum2[rt] = (sum2[rt << 1] + sum2[rt << 1 | 1]) % MOD; 23 sum3[rt] = (sum3[rt << 1] + sum3[rt << 1 | 1]) % MOD; 24 } 25 void build(int l , int r , int rt) 26 { 27 add[rt] = set[rt] = 0; 28 mul[rt] = 1; 29 if(l == r) { 30 sum1[rt] = sum2[rt] = sum3[rt] = 0; 31 return; 32 } 33 int m = (l + r) >> 1; 34 build(lson); 35 build(rson); 36 PushUp(rt); 37 } 38 void PushDown(int rt , int len) 39 { 40 if(set[rt]) { 41 set[rt << 1] = set[rt << 1 | 1 ] = the SET [RT]; 42 the Add [RT << 1 ] = the Add [RT << 1 | 1 ] = 0 ; // Note that this also delegated 43 MUL [RT << 1 ] = MUL [RT << . 1 | . 1 ] = . 1 ; 44 is LL tmp = (( SET [RT] * SET [RT])% the MOD) * SET [RT]% the MOD; 45 SUM1 [RT << . 1 ] = ((len - ( >> len . 1 )) the MOD%) * ( SET [RT]% the MOD)% MOD; 46 sum1[rt << 1 | 1] = ((len >> 1) % MOD) * (set[rt] % MOD) % MOD; 47 sum2[rt << 1] = ((len - (len >> 1)) % MOD) * ((set[rt] * set[rt]) % MOD) % MOD; 48 sum2[rt << 1 | 1] = ((len >> 1) % MOD) * ((set[rt] * set[rt]) % MOD) % MOD; 49 sum3[rt << 1] = ((len - (len >> 1)) The MOD%) * tmp% the MOD; 50 sum3 [RT << . 1 | . 1 ] = ((len >> . 1 ) the MOD%) * tmp% the MOD; 51 is SET [RT] = 0 ; 52 is } 53 is IF (MUL [RT]! = 1 ) { // this is mul [rt]! = 1, then did not pay attention to me so TLE a 54 MUL [RT << 1 ] = (MUL [RT << 1 ] * MUL [RT] )% the MOD; 55 MUL [RT << . 1 | . 1 ] = (MUL [RT << . 1 | . 1 ] * MUL [RT])% the MOD; 56 if(add[rt << 1]) //注意这个也要下放 57 add[rt << 1] = (add[rt << 1] * mul[rt]) % MOD; 58 if(add[rt << 1 | 1]) 59 add[rt << 1 | 1] = (add[rt << 1 | 1] * mul[rt]) % MOD; 60 LL tmp = (((mul[rt] * mul[rt]) % MOD * mul[rt]) % MOD); 61 sum1[rt << 1] = (sum1[rt << 1] * mul[rt]) % MOD; 62 sum1[rt << 1 | 1] = (sum1[rt << 1 | 1] * mul[rt]) % MOD; 63 sum2[rt << 1] = (sum2[rt << 1] % MOD) * ((mul[rt] * mul[rt]) % MOD) % MOD; 64 sum2[rt << 1 | 1] = (sum2[rt << 1 | 1] % MOD) * ((mul[rt] * mul[rt]) % MOD) % MOD; 65 sum3[rt << 1] = (sum3[rt << 1] % MOD) * tmp % MOD; 66 sum3[rt << 1 | 1] = (sum3[rt << 1 | 1] % MOD) * tmp % MOD; 67 mul[rt] = 1; 68 } 69 if(add[rt]) { 70 add[rt << 1] += add[rt]; //add是+= ， mul是*= 71 add[rt << 1 | 1] += add[rt]; 72 LL tmp = (add[rt] * add[rt] % MOD) * add[rt] % MOD; //注意sum3 , sum2 , sum1的先后顺序 73 sum3[rt << 1] = (sum3[rt << 1] + (tmp * (len - (len >> 1)) % MOD) + 3 * add[rt] * ((sum2[rt << 1] + sum1[rt << 1] * add[rt]) % MOD)) % MOD; 74 sum3[rt << 1 | 1] = (sum3[rt << 1 | 1] + (tmp * (len >> 1) % MOD) + 3 * add[rt] * ((sum2[rt << 1 | 1] + sum1[rt << 1 | 1] * add[rt]) % MOD)) % MOD; 75 sum2[rt << 1] = (sum2[rt << 1] + ((add[rt] * add[rt] % MOD) * (len - (len >> 1)) % MOD) + (2 * sum1[rt << 1] * add[rt] % MOD)) % MOD; 76 sum2[rt << 1 | 1] = (sum2[rt << 1 | 1] + (((add[rt] * add[rt] % MOD) * (len >> 1)) % MOD) + (2 * sum1[rt << 1 | 1] * add[rt] % MOD)) % MOD; 77 sum1[rt << 1] = (sum1[rt << 1] + (len - (len >> 1)) * add[rt]) % MOD; 78 sum1[rt <<1 | 1] = (sum1[rt << 1 | 1] + (len >> 1) * add[rt]) % MOD; 79 add[rt] = 0; 80 } 81 } 82 void update(int L , int R , int c , int ch , int l , int r , int rt) 83 { 84 if(L <= l && R >= r) { 85 if(ch == 3) { 86 set[rt] = c; 87 add[rt] = 0; 88 mul[rt] = 1; 89 sum1[rt] = ((r - l + 1) * c) % MOD; 90 sum2[rt] = ((r - l + 1) * ((c * c) % MOD)) % MOD; 91 sum3[rt] = ((r - l + 1) * (((c * c) % MOD) * c % MOD)) % MOD; 92 } else if(ch == 2) { 93 mul[rt] = (mul[rt] * c) % MOD; 94 if(add[rt]) 95 add[rt] = (add[rt] * c) % MOD; 96 sum1[rt] = (sum1[rt] * c) % MOD; 97 sum2[rt] = (sum2[rt] * (c * c % MOD)) % MOD; 98 sum3[rt] = (sum3[rt] * ((c * c % MOD) * c % MOD)) % MOD; 99 } else if(ch == 1) { 100 add[rt] += c; 101 LL tmp = (((c * c) % MOD * c) % MOD * (r - l + 1)) % MOD; //(r - l + 1) * c^3 102 sum3[rt] = (sum3[rt] + tmp + 3 * c * ((sum2[rt] + sum1[rt] * c) % MOD)) % MOD; 103 sum2[rt] = (sum2[rt] + (c * c % MOD * (r - l + 1) % MOD) + 2 * sum1[rt] * c) % MOD; 104 sum1[rt] = (sum1[rt] + (r - l + 1) * c) % MOD; 105 } 106 return; 107 } 108 PushDown(rt , r - l + 1); 109 int m = (l + r) >> 1; 110 if(L > m) 111 update(L , R , c , ch , rson); 112 else if(R <= m) 113 update(L , R , c , ch , lson); 114 else { 115 update(L , R , c , ch , lson); 116 update(L , R , c , ch , rson); 117 } 118 PushUp(rt); 119 } 120 LL query(int L , int R , int p , int l , int r , int rt) 121 { 122 if(L <= l && R >= r) { 123 if(p == 1) 124 return sum1[rt] % MOD; 125 else if(p == 2) 126 return sum2[rt] % MOD; 127 else 128 return sum3[rt] % MOD; 129 } 130 PushDown(rt , r - l + 1); 131 int m = (l + r) >> 1; 132 if(L > m) 133 return query(L , R , p , rson); 134 else if(R <= m) 135 return query(L , R , p , lson); 136 else 137 return (query(L , R , p , lson) + query(L , R , p , rson)) % MOD; 138 } 139 int main() 140 { 141 int n , m; 142 int a , b , c , ch; 143 while(~scanf("%d %d" , &n , &m)) 144 { 145 if(n == 0 && m == 0) 146 break; 147 build(1 , n , 1); 148 while(m--) { 149 scanf("%d %d %d %d" , &ch , &a , &b , &c); 150 if(ch != 4) { 151 update(a , b , c , ch , 1 , n , 1); 152 } else { 153 printf("%I64d\n" , query(a , b , c , 1 , n , 1)); 154 } 155 } 156 } 157 return 0; 158 }