Codeforces 296B Yaroslav and Strings dp+ (Getting Started

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Test instructions

Give the 2 number string s of length n, T (some positions are?)

How many kinds of filling methods make S[i]>t[i] && S[j] <t[j]

Ideas:

First find out that ans means all filling methods, ans = 10^num, num is a total number of 2 strings

Dp[0][i] Indicates the number of fill schemes with length I and for any J (1<=j<=i) satisfying s[j]<=t[j]

Dp[1][i] = S[j]==t[j]

Dp[2[i] = S[j]>=t[j]

The answer = Ans-dp[0][n]-dp[1][n] + dp[2][n];

#include <bits/stdc++.h>const int inf = 1e8;const double eps = 1e-8;const double pi = ACOs ( -1.0); const int mod = 1e9+      7;template <class t> inline BOOL Rd (T &ret) {char c; int sgn;      if (C=getchar (), c==eof) return 0; while (c!= '-' && (c< ' 0 ' | |      C> ' 9 ')) C=getchar ();      sgn= (c== '-')? -1:1;      ret= (c== '-')? 0: (c ' 0 ');      while (C=getchar (), c>= ' 0 ' &&c<= ' 9 ') ret=ret*10+ (c ' 0 ');      RET*=SGN;  return 1;      } template <class t> inline void pt (T x) {if (x <0) {Putchar ('-'); x =-X;}      if (x>9) pt (X/10);  Putchar (x%10+ ' 0 '); }using namespace Std;typedef Long long ll;typedef pair<int,int> pii;const int N = 1e5+10;int N;char s[2][n];ll dp[3] [N];void Mul (ll &x, ll y) {x = (x*y)%mod;} void Add (ll &x, ll y) {x = (x+y)%mod;}        int main () {while (cin>>n) {scanf ("%s", s[0]+1);                scanf ("%s", s[1]+1);        ll ans = 1;        for (int i = 0; i < 3; i++) dp[i][0] = 1; ll A, B, C            for (int i = 1; I <= n; i++) {if (s[0][i] = = '? ' && s[1][i] = = '? ')                {a = c = n; b = 10;            Mul (ans, 100);            } else if (s[0][i] = = '? ')                {Mul (ans, 10);                A = s[1][i]-' 0 ';                b = 1;            c = 9-a;            } else if (s[1][i] = = '? ')                {Mul (ans, 10);                c = s[0][i]-' 0 ';                b = 1;            A = 9-c;                } else {a = s[0][i]<s[1][i];                b = S[0][i]==s[1][i];            c = s[0][i]>s[1][i];            } Dp[0][i] = dp[0][i-1]* (a+b)%mod;            Dp[1][i] = Dp[1][i-1]*b%mod;        Dp[2][i] = dp[2][i-1]* (b+c)%mod;        } ans-= dp[0][n];        Ans-= dp[2][n];        Ans + = dp[1][n];        Ans%= MoD;        if (ans<0) Add (ans, mod); PT (ANS%MOD);    Puts (""); } return 0;}

Codeforces 296B Yaroslav and Strings dp+ (Getting Started