随笔分类 -  Codeforce/TopCoder/CodeChef

摘要:A: 1分钟题,往后扫一遍int a[MAXN];int vis[MAXN];int main(){ int n,m; cin>>n>>m; MEM(vis,0); for(int i = 1 ; i >a[i]; for(int j = a[i] ; j >n>>a>>b; for(int i = 0 ; i >x[i]; LL p = (x[i]*a)/b; if((p*b)%a == 0){ cout>n>>k){ int m = n/2; if(n == 1 && ... 阅读全文
posted @ 2014-04-07 13:06 Felix_F 阅读(195) 评论(0) 推荐(0) 编辑
摘要:http://www.codechef.com/NOV13还在比...我先放一部分题解吧...Uncle Johny 排序一遍struct node{ int val; int pos;}a[MAXN];int cmp(node a,node b){ return a.val >T){ while(T--){ cin>>n; for(int i = 0 ; i >a[i].val; a[i].pos = i+1; } cin>>m; ... 阅读全文
posted @ 2013-11-08 17:27 Felix_F 阅读(332) 评论(1) 推荐(0) 编辑
摘要:传送门:http://codeforces.com/contest/355A:水题,特判0int k,d;int main(){ //FIN; while(cin>>k>>d){ if(d != 0){ cout= 2){ cout>c1>>c2>>c3>>c4){ cin>>n>>m; for(int i = 0 ; i >a[i]; for(int i = 0 ; i >b[i]; LL suma = 0; for(int i = 0 ; i ... 阅读全文
posted @ 2013-10-31 20:42 Felix_F 阅读(208) 评论(0) 推荐(0) 编辑
摘要:A:sort以后求差值最小int a[100];int main(){ int n,m; cin>>n>>m; for(int i = 0 ; i >a[i]; sort(a,a+m); int mm = INF; for(int i = 0 ; i+n-1 >a>>b>>c>>d; a1 = a; b1 = b; c1 = c; d1 = d; LL p = lcm(a,c); b = b * (p/a); a = p; d = d * (p/c); c = p; if(b > d) { ... 阅读全文
posted @ 2013-08-17 10:42 Felix_F 阅读(213) 评论(0) 推荐(0) 编辑
摘要:太困了于是没做...第二天看题蘑菇题居多就只切了简单的两个...A:直接输出...int main(){ //FIN; //FOUT; int x,y; cin>>x>>y; int q = abs(abs(x) + abs(y)); if(x > 0 && y > 0) cout 0) cout 0 && y v;int main(){ //FIN; //FOUT; int n,a; cin>>n; for(int i = 0 ; i >a; v.push_back(a); ... 阅读全文
posted @ 2013-08-10 19:33 Felix_F 阅读(212) 评论(0) 推荐(0) 编辑
摘要:http://codeforces.com/contest/334A题意:1-n^2 平均分成 n 份,每份n个数,且和相同解法 : 构造矩阵1-n^2的矩阵即可int a[105][105];int main(){ int n; scanf("%d",&n); int k = n*n; int c = 1; for(int i = 0 ; i >p[i].x>>p[i].y; sort(p,p+8,cmp); for(int i = 0 ; i n且硬币数最多且这些硬币的子集不能>=n解法 : 贪心,如果非3的倍数一定是n/3+1 即全换成 阅读全文
posted @ 2013-07-28 10:48 Felix_F 阅读(354) 评论(0) 推荐(0) 编辑
摘要:A:直接判断前三项是否相等 1 int main() 2 { 3 //FIN; 4 //CHEAT; 5 int n; 6 cin>>n; 7 getchar(); 8 char a[4005]; 9 gets(a);10 int len = strlen(a);11 int cnt = 0;12 for(int i = 0 ; i >n>>k;12 for(int i = 1 ; i >a[i];15 sum[i] = sum[i-1] + a[i];16 }17 int ... 阅读全文
posted @ 2013-07-26 00:01 Felix_F 阅读(352) 评论(1) 推荐(0) 编辑

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