HIT summer training Contest 11 / A step number
2011-08-05 19:16
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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
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; //456 654 x/=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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