1.
hdu1698
http://acm.hdu.edu.cn/showproblem.php?pid=1698
1 /* 2 x y k 3 x~y的值变为k 4 */ 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cmath> 8 #include <cstring> 9 #include <time.h> 10 #include <string> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <stack> 15 #include <queue> 16 #include <vector> 17 #include <bitset> 18 #include <ext/rope> 19 #include <algorithm> 20 #include <iostream> 21 using namespace std; 22 #define ll long long 23 #define minv 1e-6 24 #define inf 1e9 25 #define pi 3.1415926536 26 #define E 2.7182818284 27 const ll mod=1e9+7;//998244353 28 const int maxn=1e5+10; 29 30 int tag[maxn<<2],sum[maxn<<2]; 31 32 void push_down(int index,int len) 33 { 34 tag[index<<1]=tag[index<<1|1]=tag[index]; 35 sum[index<<1]=((len+1)>>1)*tag[index]; 36 sum[index<<1|1]=(len>>1)*tag[index]; 37 tag[index]=0; 38 } 39 40 void build(int index,int l,int r) 41 { 42 tag[index]=0; 43 if (l==r) 44 sum[index]=1; 45 else 46 { 47 int m=(l+r)>>1; 48 build(index<<1,l,m); 49 build(index<<1|1,m+1,r); 50 sum[index]=sum[index<<1]+sum[index<<1|1]; 51 } 52 } 53 54 void update(int index,int l,int r,int x,int y,int k) 55 { 56 if (x<=l && r<=y) 57 { 58 tag[index]=k; 59 sum[index]=(r-l+1)*k; 60 return; 61 } 62 if (tag[index]!=0) 63 push_down(index,r-l+1); 64 int m=(l+r)>>1; 65 if (x<=m) 66 update(index<<1,l,m,x,y,k); 67 if (m<y) 68 update(index<<1|1,m+1,r,x,y,k); 69 sum[index]=sum[index<<1]+sum[index<<1|1]; 70 } 71 72 int query(int index,int l,int r,int s,int t) 73 { 74 if (s<=l && r<=t) 75 return sum[index]; 76 if (r<s || l>t) 77 return 0; 78 if (tag[index]!=0) 79 push_down(index,r-l+1); 80 int m=(l+r)>>1; 81 return query(index<<1,l,m,s,t)+query(index<<1|1,m+1,r,s,t); 82 } 83 84 int main() 85 { 86 int t,T,n,q,x,y,k; 87 scanf("%d",&t); 88 for (T=1;T<=t;T++) 89 { 90 scanf("%d",&n); 91 build(1,1,n); 92 scanf("%d",&q); 93 while (q--) 94 { 95 scanf("%d%d%d",&x,&y,&k); 96 update(1,1,n,x,y,k); 97 } 98 printf("Case %d: The total value of the hook is %d.\n",T,query(1,1,n,1,n)); 99 } 100 return 0; 101 }
2.
https://www.luogu.org/problemnew/show/P3372
1 /* 2 第八个测试数据有问题(已看) 3 4 操作1： 格式：1 x y k 含义：将区间[x,y]内每个数加上k 5 6 操作2： 格式：2 x y 含义：输出区间[x,y]内每个数的和 7 */ 8 #include <cstdio> 9 #include <cstdlib> 10 #include <cmath> 11 #include <cstring> 12 #include <time.h> 13 #include <string> 14 #include <set> 15 #include <map> 16 #include <list> 17 #include <stack> 18 #include <queue> 19 #include <vector> 20 #include <bitset> 21 #include <ext/rope> 22 #include <algorithm> 23 #include <iostream> 24 using namespace std; 25 #define ll long long 26 #define minv 1e-6 27 #define inf 1e9 28 #define pi 3.1415926536 29 #define E 2.7182818284 30 const ll mod=1e9+7;//998244353 31 const int maxn=1e5+10; 32 33 ll sum[maxn<<2],tag[maxn<<2]; 34 35 void build(int index,int l,int r) 36 { 37 tag[index]=0; 38 if (l==r) 39 { 40 scanf("%lld",&sum[index]); 41 sum[index]=sum[index]%mod; 42 } 43 else 44 { 45 int m=(l+r)>>1; 46 build(index<<1,l,m); 47 build(index<<1|1,m+1,r); 48 sum[index]=(sum[index<<1]+sum[index<<1|1])%mod; 49 } 50 } 51 52 void pushdown(int index,ll len) 53 { 54 sum[index<<1]=(tag[index]*((len+1)>>1) +sum[index<<1])%mod; 55 sum[index<<1|1]=(tag[index]*(len>>1) +sum[index<<1|1])%mod; 56 tag[index<<1]=(tag[index]+ tag[index<<1])%mod; 57 tag[index<<1|1]=(tag[index]+ tag[index<<1|1])%mod; 58 tag[index]=0; 59 } 60 61 void update(int index,int l,int r,int x,int y,ll k) 62 { 63 if (x<=l && r<=y) 64 { 65 sum[index]=(k*(r-l+1) +sum[index])%mod; 66 tag[index]=(k +tag[index])%mod; 67 return; 68 } 69 int m=(l+r)>>1; 70 if (tag[index]!=0) 71 pushdown(index,r-l+1); 72 if (x<=m) 73 update(index<<1,l,m,x,y,k); 74 if (m<y) 75 update(index<<1|1,m+1,r,x,y,k); 76 sum[index]=(sum[index<<1]+sum[index<<1|1])%mod; 77 } 78 79 ll query(int index,int l,int r,int x,int y) 80 { 81 if (x<=l && r<=y) 82 return sum[index]; 83 if (r<x || l>y) 84 return 0; 85 if (tag[index]!=0) 86 pushdown(index,r-l+1); 87 int m=(l+r)>>1; 88 return (query(index<<1,l,m,x,y)+query(index<<1|1,m+1,r,x,y))%mod; 89 } 90 91 int main() 92 { 93 int n,m,mode,x,y; 94 ll k; 95 scanf("%d%d",&n,&m); 96 build(1,1,n); 97 while (m--) 98 { 99 scanf("%d",&mode); 100 if (mode==1) 101 { 102 scanf("%d%d%lld",&x,&y,&k); 103 update(1,1,n,x,y,k%mod); 104 } 105 else 106 { 107 scanf("%d%d",&x,&y); 108 printf("%lld\n",query(1,1,n,x,y)); 109 } 110 } 111 return 0; 112 } 113 /* 114 10 115 463 793 740 374 330 772 681 116 5 8 39 117 5 8 118 3 6 3 119 5 8 90 120 1 5 21 121 3 8 122 3 8 17 123 4 7 52 124 2 6 125 2 7 41 126 5 127 2 3 4 5 128 1 3 -1 129 1 3 1 130 2 4 131 100 132 2 3 4 5 6 7 8 9 10 133 1 10 134 2 7 1 135 1 10 136 3 8 137 138 */
3.
https://www.luogu.org/problemnew/show/P3373
1 /* 2 1.将某区间每一个数乘上x 3 4 2.将某区间每一个数加上x 5 6 3.求出某区间每一个数的和 7 */ 8 #include <cstdio> 9 #include <cstdlib> 10 #include <cmath> 11 #include <cstring> 12 #include <time.h> 13 #include <string> 14 #include <set> 15 #include <map> 16 #include <list> 17 #include <stack> 18 #include <queue> 19 #include <vector> 20 #include <bitset> 21 #include <ext/rope> 22 #include <algorithm> 23 #include <iostream> 24 using namespace std; 25 #define ll long long 26 #define minv 1e-6 27 #define inf 1e9 28 #define pi 3.1415926536 29 #define E 2.7182818284 30 //const ll mod=1e9+7;//998244353 31 const int maxn=1e5+10; 32 33 ll mod; 34 ll add[maxn<<2],mul[maxn<<2],sum[maxn<<2]; 35 int mode; 36 37 void build(int index,int l,int r) 38 { 39 add[index]=0; 40 mul[index]=1; 41 if (l==r) 42 { 43 scanf("%lld",&sum[index]); 44 sum[index]=sum[index]%mod; 45 } 46 else 47 { 48 int m=(l+r)>>1; 49 build(index<<1,l,m); 50 build(index<<1|1,m+1,r); 51 sum[index]=(sum[index<<1]+sum[index<<1|1])%mod; 52 } 53 } 54 55 void pushdown(int index,ll len) 56 { 57 //(x *y+z)*s+t = x*(ys) + z*s+t 58 add[index<<1]=(add[index<<1]*mul[index]+add[index])%mod; 59 mul[index<<1]=mul[index]*mul[index<<1]%mod; 60 sum[index<<1]=(sum[index<<1]*mul[index]+add[index]*((len+1)>>1))%mod; 61 62 add[index<<1|1]=(add[index<<1|1]*mul[index]+add[index])%mod; 63 mul[index<<1|1]=mul[index]*mul[index<<1|1]%mod; 64 sum[index<<1|1]=(sum[index<<1|1]*mul[index]+add[index]*(len>>1))%mod; 65 66 add[index]=0; 67 mul[index]=1; 68 } 69 70 void update(int index,int l,int r,int x,int y,ll k) 71 { 72 if (x<=l && r<=y) 73 { 74 if (mode==1) 75 { 76 //(x*y+z)*k 77 sum[index]=sum[index]*k%mod; 78 add[index]=add[index]*k%mod; 79 mul[index]=mul[index]*k%mod; 80 } 81 else 82 { 83 //(x*y+z)+k 84 sum[index]=(sum[index]+k*(r-l+1))%mod; 85 add[index]=(add[index]+k)%mod; 86 } 87 return; 88 } 89 pushdown(index,r-l+1); 90 int m=(l+r)>>1; 91 if (x<=m) 92 update(index<<1,l,m,x,y,k); 93 if (m<y) 94 update(index<<1|1,m+1,r,x,y,k); 95 sum[index]=(sum[index<<1]+sum[index<<1|1])%mod; 96 } 97 98 ll query(int index,int l,int r,int x,int y) 99 { 100 if (x<=l && r<=y) 101 return sum[index]; 102 if (x>r || y<l) 103 return 0; 104 pushdown(index,r-l+1); 105 int m=(l+r)>>1; 106 return (query(index<<1,l,m,x,y)+query(index<<1|1,m+1,r,x,y))%mod; 107 } 108 109 int main() 110 { 111 int n,q,x,y; 112 ll k; 113 scanf("%d%d%lld",&n,&q,&mod); 114 build(1,1,n); 115 while (q--) 116 { 117 scanf("%d%d%d",&mode,&x,&y); 118 if (mode!=3) 119 { 120 scanf("%lld",&k); 121 update(1,1,n,x,y,k%mod); 122 } 123 else 124 printf("%lld\n",query(1,1,n,x,y)); 125 } 126 return 0; 127 } 128 /* 129 5 100 1000000 130 1 2 3 4 5 131 1 1 5 2 132 2 1 5 3 133 3 1 3 134 135 */