分析
区间加，区间和，区间最小值，区间整除
跟区间开方的性质差不多，一次一次除下去，区间中的数会越来越接近
当区间中的数接近时，他们减少的数会相等，就转换成了区间减
代码
1 /***************************** 2 User:Mandy.H.Y 3 Language:c++ 4 Problem:market 5 *****************************/ 6 // 7 #include<bits/stdc++.h> 8 #define lson l,mid,k << 1 9 #define rson mid + 1,r,k << 1 | 1 10 #define Max(x,y) ((x) > (y) ? (x) : (y)) 11 #define Min(x,y) ((x) < (y) ? (x) : (y)) 12 13 using namespace std; 14 15 const int maxn = 1e5 + 5; 16 17 int n,q; 18 long long sum[maxn << 2]; 19 int mi[maxn << 2],ma[maxn << 2],lazy[maxn << 2]; 20 int L,R,C,D; 21 22 template<class T>inline void read(T &x){ 23 x = 0;bool flag = 0;char ch = getchar(); 24 while(!isdigit(ch)) flag |= ch == '-',ch = getchar(); 25 while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar(); 26 if(flag) x = -x; 27 } 28 29 template<class T>void putch(const T x){ 30 if(x > 9) putch(x / 10); 31 putchar(x % 10 | 48); 32 } 33 34 template<class T>void put(const T x){ 35 if(x < 0) putchar('-'),putch(-x); 36 else putch(x); 37 } 38 39 void file(){ 40 freopen("market.in","r",stdin); 41 freopen("market.out","w",stdout); 42 } 43 44 void pushup(int k){ 45 int ls = k << 1; 46 int rs = k << 1 | 1; 47 sum[k] = sum[ls] + sum[rs]; 48 mi[k] = Min(mi[ls],mi[rs]); 49 ma[k] = Max(ma[ls],ma[rs]); 50 } 51 52 void buildtree(int l,int r,int k){ 53 if(l == r){ 54 read(mi[k]); 55 sum[k] = ma[k] = mi[k]; 56 return ; 57 } 58 int mid = (l + r) >> 1; 59 buildtree(lson); 60 buildtree(rson); 61 pushup(k); 62 } 63 64 void readdata(){ 65 read(n);read(q); 66 buildtree(1,n,1); 67 } 68 69 void pushdown(int l,int r,int k){ 70 if(!lazy[k]) return; 71 int ls = k << 1; 72 int rs = ls | 1; 73 int mid = (l + r) >> 1; 74 sum[ls] += (long long)lazy[k] * (mid - l + 1); 75 sum[rs] += (long long)lazy[k] * (r - mid); 76 mi[ls] += lazy[k];ma[ls] += lazy[k]; 77 mi[rs] += lazy[k];ma[rs] += lazy[k]; 78 lazy[ls] += lazy[k]; 79 lazy[rs] += lazy[k]; 80 lazy[k] = 0; 81 } 82 83 void add(int l,int r,int k){ 84 if(L <= l && r <= R){ 85 sum[k] += (long long)C * (r - l + 1); 86 mi[k] += C; 87 ma[k] += C; 88 lazy[k] += C; 89 return; 90 } 91 pushdown(l,r,k); 92 int mid = (l + r) >> 1; 93 if(L <= mid) add(lson); 94 if(R > mid) add(rson); 95 pushup(k); 96 } 97 98 void div(int l,int r,int k){//如果区间最大值与最小值的下降的值都一样，那么区间下降的值一样 99 if(L <= l && r <= R){ 100 int tmp1 = ma[k] - (int)floor((double)ma[k]/(double)D); 101 int tmp2 = mi[k] - (int)floor((double)mi[k]/(double)D); 102 if(tmp1 == tmp2){ 103 sum[k] -= (long long)tmp1 * (r - l + 1); 104 mi[k] -= tmp1; 105 ma[k] -= tmp1; 106 lazy[k] -= tmp1; 107 return; 108 } 109 } 110 pushdown(l,r,k); 111 int mid = (l + r) >> 1; 112 if(L <= mid) div(lson); 113 if(R > mid) div(rson); 114 pushup(k); 115 } 116 117 long long get_sum(int l,int r,int k){ 118 if(L <= l && r <= R) return sum[k]; 119 pushdown(l,r,k); 120 int mid = (l + r) >> 1; 121 long long ans = 0; 122 if(L <= mid) ans += get_sum(lson); 123 if(R > mid) ans += get_sum(rson); 124 pushup(k); 125 return ans; 126 } 127 128 int get_min(int l,int r,int k){ 129 if(L <= l && r <= R) return mi[k]; 130 pushdown(l,r,k); 131 int mid = (l + r) >> 1; 132 int ans1 = 2e9 + 5; 133 int ans2 = 2e9 + 5; 134 if(L <= mid) ans1 = get_min(lson); 135 if(R > mid) ans2 = get_min(rson); 136 pushup(k); 137 return Min(ans1,ans2); 138 } 139 140 void work(){ 141 while(q --){ 142 int opt; 143 read(opt); 144 switch(opt){ 145 case 1:{ 146 read(L);read(R);read(C); 147 L++,R++;//注意本题给出的标号从0开始 148 add(1,n,1); 149 break; 150 } 151 case 2:{ 152 read(L);read(R);read(D); 153 L++,R++; 154 div(1,n,1); 155 break; 156 } 157 case 3:{ 158 read(L);read(R);L++,R++; 159 int ans = get_min(1,n,1); 160 put(ans);puts(""); 161 break; 162 } 163 case 4:{ 164 read(L);read(R); 165 L++,R++; 166 long long ans = get_sum(1,n,1); 167 put(ans);puts(""); 168 break; 169 } 170 } 171 } 172 } 173 174 int main(){ 175 // file(); 176 readdata(); 177 work(); 178 return 0; 179 }