HIT summer training Contest 11 / A step number
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目录
题目
这题我不知道为什么我想了n久,就是没开敲。傻逼般地觉得n<= 2^31会超时,果真傻逼。
后来比赛结束后敲了一下,程序效率很高啊,尼玛才反应过来这是按位的,10位每个位搜......,尼玛超个鬼。。。。 应该说搜的是一个特殊的解答树,左边可以尽量展开0-9子节点,越往右结点急剧减小,在输入 a=1 b=999999999 时,只搜了43758次。
反思:记得看刘汝佳的白书的时候他是有提到这种情况的,以后分析要仔细,不要怕超不敢做,好几次这样了。。。
这代码写得很乱
#include<cstdio>#include<cstring>#include<iostream>using namespace std;int sum;void solve(int isLim, int num[], int preMax, int cur, int n, int ans){int curLim;if(isLim) curLim = num[cur];else curLim = 9;if(cur == n-1){sum+=(curLim-preMax+1);//printf("%d %d\n", ans, (curLim-preMax+1));}else{if(curLim>=preMax){for(int pre=preMax; pre<=curLim; pre++)solve((isLim&&(pre==curLim)), num, pre, cur+1, n, ans*10+pre);}}}int getnum(int x, int num[]){int buf[20];int i;for(i=0; x; i++){buf[i] = x%10;//456 654x/=10;}for(int j=0; j<i; j++){num[j] = buf[i-j-1];}return i;}int main(){int a, b;//1<<31 = 2147483648其中有一个1,所以无影响,int32也可以int num[20]; while(scanf("%d%d", &a, &b) != EOF){if(a<0) a = (1<<31)-1;if(b<0) b = (1<<31)-1;int n = getnum(a-(a==1?0:1), num);solve(1, num, 0, 0, n, 0);int ans = sum;n = getnum(b, num);sum=0;solve(1, num, 0, 0, n, 0);ans = sum-ans;if(a == 1) ans+=1;printf("%d\n", ans);sum=0;}}
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