D2 - Submarine in the Rybinsk Sea (hard edition)
千万注意，是%=，而不是%
(sum+=value[i][len[i]]*g[j])%=mod;
与
(sum+=value[i][len[i]]*g[j])%mod; //wrong
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <iostream> 8 using namespace std; 9 #define ll long long 10 11 const double eps=1e-8; 12 const ll inf=1e9; 13 const ll mod=998244353; 14 const int maxn=1e5+10; 15 16 ll len[maxn],a[maxn],value[maxn][21],ext[maxn][21],ch[21],g[21]; 17 18 ll cal(ll d) 19 { 20 ll tot=0; 21 while (d) 22 { 23 tot++; 24 d/=10; 25 } 26 return tot; 27 } 28 29 int main() 30 { 31 ll i,j,k,n,d,sum=0,add; 32 ch[0]=1; 33 for (i=1;i<=20;i++) 34 ch[i]=ch[i-1]*10%mod;///10^i 35 36 scanf("%lld",&n); 37 for (i=1;i<=n;i++) 38 { 39 scanf("%lld",&a[i]); 40 d=a[i]; 41 len[i]=cal(d); 42 g[len[i]]++; 43 44 j=1; 45 add=0; 46 while (d) 47 { 48 (add+=d%10*ch[j*2-2])%=mod; 49 value[i][j]=add*11%mod; 50 d/=10; 51 ext[i][j]=d*ch[j*2]*2%mod; 52 j++; 53 } 54 // printf("z"); 55 } 56 57 for (i=1;i<=n;i++) 58 { 59 k=len[i]; 60 for (j=1;j<=10;j++) 61 if (k<=j) 62 (sum+=value[i][len[i]]*g[j])%=mod; 63 else 64 { 65 (sum+=(value[i][j]+ext[i][j])*g[j])%=mod; 66 } 67 } 68 69 printf("%lld",sum);///why 70 return 0; 71 } 72 /* 73 5 74 1000000000 1000000000 1000000000 1000000000 1000000000 75 76 in codeblocks, answer is wrong? 77 ³¢ÊÔÌá½»£¿ 78 */
套路单调队列题
从上到下记录a*1，然后从左往右记录a*b(若干个a*1)
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <iostream> 8 using namespace std; 9 #define ll long long 10 11 const double eps=1e-8; 12 const ll inf=1e9; 13 const ll mod=1e9+7; 14 const int maxn=3e3+10; 15 16 ll h[maxn][maxn],pos[maxn]; 17 ll col[maxn][maxn]; 18 19 int main() 20 { 21 ll sum=0; 22 ll n,m,a,b,i,j; 23 ll g0,x,y,z; 24 ll head,tail; 25 scanf("%lld%lld%lld%lld",&n,&m,&a,&b); 26 scanf("%lld%lld%lld%lld",&g0,&x,&y,&z); 27 for (i=1;i<=n;i++) 28 for (j=1;j<=m;j++) 29 { 30 h[i][j]=g0; 31 g0=(1ll*g0*x+y)%z; 32 } 33 34 35 for (j=1;j<=m;j++) 36 { 37 head=1,tail=0; 38 for (i=1;i<=n;i++) 39 { 40 while (head<=tail && pos[head]<=i-a) 41 head++; 42 while (head<=tail && h[pos[tail]][j]>=h[i][j]) 43 tail--; 44 tail++; 45 pos[tail]=i; 46 col[i][j]=h[pos[head]][j]; 47 } 48 } 49 50 for (i=a;i<=n;i++) 51 { 52 head=1,tail=0; 53 for (j=1;j<=m;j++) 54 { 55 while (head<=tail && pos[head]<=j-b) 56 head++; 57 while (head<=tail && col[i][pos[tail]]>=col[i][j]) 58 tail--; 59 tail++; 60 pos[tail]=j; 61 if (j>=b) 62 sum+=col[i][pos[head]]; 63 } 64 } 65 printf("%lld",sum); 66 return 0; 67 } 68 /* 69 3 4 2 1 70 1000000000 1000000000 1 234142 71 */