代码搜索：continue

找到约 10,000 项符合「continue」的源代码

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热门代码： main.c GPIO I2C SPI UART timer interrupt ADC PWM LED
www.eeworm.com/read/288479/8630031

m stixdemo.m

%STIXDEMO Demonstrate various stixbox routines. % Anders Holtsberg, 25-11-94 % Copyright (c) Anders Holtsberg hold off subplot(111) echo on clc % ========================================
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www.eeworm.com/read/431812/8652427

cpp pku2190.cpp

#include "stdio.h" #include "string.h" int main() { char s[11]; int i,N; int total; int k,flag=-1; while(scanf("%s",s)!=-1) { total=0; flag=-1; k=0; for(i=0;i
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www.eeworm.com/read/431812/8652475

cpp pku2139.cpp

#include int main() { int i, j, k, N, M, n, min; int mb[1000]; int dis[301][301]; scanf("%d%d", &N, &M); for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { d
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www.eeworm.com/read/431812/8652668

cpp pku2613.cpp

#include "stdio.h" int main() { int p[10000],i; int a,b,c,d; double out; while(scanf("%d%d%d%d",&a,&b,&c,&d)!=-1) { out=1.0; for(i=0;ib;i--)p[i]++;
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www.eeworm.com/read/431812/8652717

cpp pku2552.cpp

#include int main() { int status[35000], man[4000]; int i, j, p; for(i=0; i
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www.eeworm.com/read/431812/8652753

cpp pku2573.cpp

#include #include int B[1001], W[5000]; int N; int main() { int i, ans, e, a, b, c, d, k; while (scanf("%d", &N) != -1) { for (i = 0; i < N; i++) { scanf(
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www.eeworm.com/read/431812/8652940

cpp pku2363.cpp

#include #include int main() { int t, n, a, b, c, min = 10000; scanf("%d", &t); while (t--) { min = 10000; scanf("%d", &n); if (n == 0) { printf("0\n");
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www.eeworm.com/read/431812/8653029

cpp pku2704.cpp

#include "stdio.h" int main() { long long way[34][34]; char num[34][34]; int n,i,j,step; while(1) { scanf("%d",&n); if(n==-1) break; for(i=0;i
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www.eeworm.com/read/431812/8653033

cpp pku2780.cpp

#include #include typedef struct { int x, y; }Point; Point p[1001], d[1001]; int N; int Abs(int x) { return (x > 0) ? x : -x; } int sgn(int x) { if (x >
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www.eeworm.com/read/431812/8653484

cpp pku1411.cpp

#include #define Max 100001 int prime[Max]; void pre() { int i, j; memset(prime, -1, sizeof(prime)); for (i = 2; i * i < Max; i++) { if (prime[i] == 0) { continue;
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