hihocoder Challenge 29 B.快速乘法

这题的题解和我写的有一拼，异常简洁，爆炸。

这题思路dp
dp[i][j] 表示的是讨论到第n位，并比原数的前n位多了 j∗2i
显然j只能取0,1,毕竟2进制嘛
之后转移就好了,注意下面两个重要状态
dp[i][1]=dp[i][0]+2
dp[i][0]=dp[i][1]+2

#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <queue>using namespace std;const int N = 1e6+5;const int MOD = 998244353;const int INF = 0x3f3f3f3f;char s[N];int dp[N][2];void gmin(int &a, int b) {    if(a > b) a = b;}int main() {    while(~scanf("%s", s)) {        int len = strlen(s);        memset(dp, INF, sizeof(dp));        for(int i = len+1; i >= 2; --i) s[i] = s[i-2];     //  for(int i = 2; i < len+2; ++i) printf("%c", s[i]); printf("\n");        s[1] = '0';  s[len+2] = '0';    //  for(int i = 2; i <= len+1; ++i) printf("%c", s[i]); printf("\n");        dp[0][0] = 0;        for(int i = 1; i <= len+2; ++i) {            for(int j = 0; j < 2; ++j) {            //  if(dp[i-1][j] == INF) continue;                if(j == 0) {                    if(s[i] == '0') {                        gmin(dp[i][0], dp[i-1][0]);                        gmin(dp[i][1], dp[i-1][0] + 2);                    } else {                        gmin(dp[i][0], dp[i-1][0] + 2);                    }                }else {                    if(s[i] == '0') {                        gmin(dp[i][1], dp[i-1][1] + 2);                    }else {                        gmin(dp[i][0], dp[i-1][1] + 2);                        gmin(dp[i][1], dp[i-1][1]);                    }                }                           }        //  printf("%d:%d %d\n", i, dp[i][0], dp[i][1]);        }        printf("%d\n", dp[len+2][0]-1);    }    return 0;}